how to solve equations in square root

How to Solve Quadratic Equations using the Square Root Method

This is the “best” method whenever the quadratic equation only contains {x^2} terms. That implies no presence of any x term being raised to the first power somewhere in the equation.

Key Strategy in Solving Quadratic Equations using the Square Root Method

The general approach is to collect all {x^2} terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of x . Always attach the \pm symbol when you get the square root of the constant.

Examples of How to Solve Quadratic Equations by Square Root Method

Example 1 : Solve the quadratic equation below using the Square Root Method.

I will isolate the only {x^2} term on the left side by adding both sides by + 1 . Then solve the values of x by taking the square roots of both sides of the equation. As I mentioned before, we need to attach the plus or minus symbol to the square root of the constant.

x is equal to positive 5 or x is equal to negative 5

So I have x = 5 and x = - \,5 as final answers since both of these values satisfy the original quadratic equation. I will leave it to you to verify.

Example 2 : Solve the quadratic equation below using the Square Root Method.

This problem is very similar to the previous example. The only difference is that after I have separated the {x^2} term and the constant in the opposite sides of the equation, I need to divide the equation by the coefficient of the squared term before taking the square roots of both sides.

x is equal to positive 4 or x is equal to negative 4

The final answers are x = 4 and x = - \,4 .

Example 3 : Solve the quadratic equation below using the Square Root Method.

I can see that I have two {x^2} terms, one on each side of the equation. My approach is to collect all the squared terms of x to the left side, and combine all the constants to the right side. Then solve for x as usual, just like in Examples 1 and 2.

x is equal to 3 or x is equal to negative 3

The solutions to this quadratic formula are x = 3 and x = - \,3 .

Example 4 : Solve the quadratic equation below using the Square Root Method.

The two parentheses should not bother you at all. The fact remains that all variables come in the squared form, which is what we want. This problem is perfectly solvable using the square root method.

So my first step is to eliminate both of the parentheses by applying the distributive property of multiplication. Once they are gone, I can easily combine like terms. Keep the {x^2} terms to the left, and constants to the right. Finally, apply square root operation in both sides and we’re done!

Not too bad, right?

Example 5 : Solve the quadratic equation below using the Square Root Method.

Since the x -term is being raised to the second power twice, that means, I need to perform two square root operations in order to solve for x .

The first step is to have something like this: ( ) 2 = constant . This allows me to get rid of the exponent of the parenthesis on the first application of square root operation.

After doing so, what remains is the “stuff” inside the parenthesis which has an {x^2} term. Well, this is great since I already know how to handle it just like the previous examples.

There’s an x -squared term left after the first application of square root.

Now we have to break up {x^2} = \pm \,6 + 10 into two cases because of the “plus” or “minus” in 6 .

Solve the first case where 6 is positive .

Solve the second case where 6 is negative .

The solutions to this quadratic equations are x = 4 , x = - \,4 , x = 2 , and x = - \,2 . Yes, we have four values of x that can satisfy the original quadratic equation.

Example 6 : Solve the quadratic equation below using the Square Root Method.

Example 7 : Solve the quadratic equation below using the Square Root Method.

Unit 10: Lesson 2

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9.2: Solve Quadratic Equations Using the Square Root Property

Learning Objectives

By the end of this section, you will be able to:

Before you get started, take this readiness quiz.

A quadratic equation is an equation of the form $a x^{2}+b x+c=0$ , where $a≠0$. Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form $ax^{2}$. We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the Form $ax^{2}=k$ using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation $x^{2}=9$.

$x^{2}=9$

Put the equation in standard form.

$x^{2}-9=0$

Factor the difference of squares.

$(x-3)(x+3)=0$

Use the Zero Produce Property.

$x-3=0 \quad x-3=0$

Solve each equation.

$x=3 \quad x=-3$

We can easily use factoring to find the solutions of similar equations, like $x^{2}=16$ and $x^{2}=25$, because $16$ and $25$ are perfect squares. In each case, we would get two solutions, $x=4, x=-4$ and $x=5, x=-5$

But what happens when we have an equation like $x^{2}=7$? Since $7$ is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since $169$ is the square of $13$, we can also say that $13$ is a square root of $169$. Also, $(-13)^{2}=169$, so $−13$ is also a square root of $169$. Therefore, both $13$ and $−13$ are square roots of $169$. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

If $n^{2}=m$, then $n$ is a square root of $m$.

Since these equations are all of the form $x^{2}=k$, the square root definition tells us the solutions are the two square roots of $k$. This leads to the Square Root Property .

Definition $\PageIndex{1}$

Square Root Property

If $x^{2}=k$, then

$x=\sqrt{k} \quad$ or $\quad x=-\sqrt{k} \quad$ or $\quad x=\pm \sqrt{k}$

Notice that the Square Root Property gives two solutions to an equation of the form $x^{2}=k$, the principal square root of $k$ and its opposite. We could also write the solution as $x=\pm \sqrt{k}$. We read this as $x$ equals positive or negative the square root of $k$.

Now we will solve the equation $x^{2}=9$ again, this time using the Square Root Property.

$\begin{aligned} &x^{2} =9 \\ \text { Use the Square Root Property. } \quad& x=\pm \sqrt{9} \\& x =\pm 3 \end{aligned}$

So $x=3$ or $x=-3$

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation $x^{2}=7$.

$x^{2}=7$

Use the Square Root Property. $x=\sqrt{7}, \quad x=-\sqrt{7}$

We cannot simplify $\sqrt{7}$, so we leave the answer as a radical.

Example $\PageIndex{1}$ How to Solve a Quadratic Equation of the form $ax^{2}-k$ Using the Square Root Property

Solve: $x^{2}-50=0$.

Exercise $\PageIndex{1}$

Solve: $x^{2}-48=0$.

$x=4 \sqrt{3}, x=-4 \sqrt{3}$

Exercise $\PageIndex{2}$

Solve: $y^{2}-27=0$.

$y=3 \sqrt{3}, y=-3 \sqrt{3}$

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a Quadratic Equation Using the Square Root Property

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient $3$ before using the Square Root Property.

Example $\PageIndex{2}$

Solve: $3 z^{2}=108$.

Exercise $\PageIndex{3}$

Solve: $2x^{2}=98$.

$x=7, x=-7$

Exercise $\PageIndex{4}$

Solve: $5m^{2}=80$.

$m=4, m=-4$

The Square Root Property states ‘If $x^{2}=k$,’ What will happen if $k<0$? This will be the case in the next example.

Example $\PageIndex{3}$

Solve: $x^{2}+72=0$.

Exercise $\PageIndex{5}$

Solve: $c^{2}+12=0$.

$c=2 \sqrt{3} i, \quad c=-2 \sqrt{3} i$

Exercise $\PageIndex{6}$

Solve: $q^{2}+24=0$.

$c=2 \sqrt{6} i, \quad c=-2 \sqrt{6} i$

Our method also works when fractions occur in the equation, we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Example $\PageIndex{4}$

Solve: $\frac{2}{3} u^{2}+5=17$.

Exercise $\PageIndex{7}$

Solve: $\frac{1}{2} x^{2}+4=24$.

$x=2 \sqrt{10}, x=-2 \sqrt{10}$

Exercise $\PageIndex{8}$

Solve: $\frac{3}{4} y^{2}-3=18$.

$y=2 \sqrt{7}, y=-2 \sqrt{7}$

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .

Example $\PageIndex{5}$

Solve: $2 x^{2}-8=41$.

Exercise $\PageIndex{9}$

Solve: $5 r^{2}-2=34$.

$r=\frac{6 \sqrt{5}}{5}, \quad r=-\frac{6 \sqrt{5}}{5}$

Exercise $\PageIndex{10}$

Solve: $3 t^{2}+6=70$.

$t=\frac{8 \sqrt{3}}{3}, \quad t=-\frac{8 \sqrt{3}}{3}$

Solve Quadratic Equation of the Form $a(x-h)^{2}=k$ Using the Square Root Property

We can use the Square Root Property to solve an equation of the form $a(x-h)^{2}=k$ as well. Notice that the quadratic term, $x$, in the original form $ax^{2}=k$ is replaced with $(x-h)$.

$On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.$

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of $a$, then the Square Root Property can be used on $(x-h)^{2}$.

Example $\PageIndex{6}$

Solve: $4(y-7)^{2}=48$.

Exercise $\PageIndex{11}$

Solve: $3(a-3)^{2}=54$.

$a=3+3 \sqrt{2}, \quad a=3-3 \sqrt{2}$

Exercise $\PageIndex{12}$

Solve: $2(b+2)^{2}=80$.

$b=-2+2 \sqrt{10}, \quad b=-2-2 \sqrt{10}$

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example $\PageIndex{7}$

Solve: $\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}$.

$\left(x-\frac{1}{3}\right)^{2}=\frac{5}{9}$

Use the Square Root Property.

$x-\frac{1}{3}=\pm \sqrt{\frac{5}{9}}$

Rewrite the radical as a fraction of square roots.

$x-\frac{1}{3}=\pm \frac{\sqrt{5}}{\sqrt{9}}$

$x-\frac{1}{3}=\pm \frac{\sqrt{5}}{3}$

Solve for $x$.

$x=\frac{1}{3} \pm \frac{\sqrt{5}}{3}$

Rewrite to show two solutions.

$x=\frac{1}{3}+\frac{\sqrt{5}}{3}, x=\frac{1}{3}-\frac{\sqrt{5}}{3}$

We leave the check for you.

Exercise $\PageIndex{13}$

Solve: $\left(x-\frac{1}{2}\right)^{2}=\frac{5}{4}$.

$x=\frac{1}{2}+\frac{\sqrt{5}}{2}, x=\frac{1}{2}-\frac{\sqrt{5}}{2}$

Exercise $\PageIndex{14}$

Solve: $\left(y+\frac{3}{4}\right)^{2}=\frac{7}{16}$.

$y=-\frac{3}{4}+\frac{\sqrt{7}}{4}, y=-\frac{3}{4}-\frac{\sqrt{7}}{4}$

We will start the solution to the next example by isolating the binomial term.

Example $\PageIndex{8}$

Solve: $2(x-2)^{2}+3=57$.

$2(x-2)^{2}+3=57$

Subtract $3$ from both sides to isolate the binomial term.

$2(x-2)^{2}=54$

Divide both sides by $2$.

$x=2+3 \sqrt{3}, x=2-3 \sqrt{3}$

Exercise $\PageIndex{15}$

Solve: $5(a-5)^{2}+4=104$.

$a=5+2 \sqrt{5}, a=5-2 \sqrt{5}$

Exercise $\PageIndex{16}$

Solve: $3(b+3)^{2}-8=88$.

$b=-3+4 \sqrt{2}, \quad b=-3-4 \sqrt{2}$

Sometimes the solutions are complex numbers.

Example $\PageIndex{9}$

Solve: $(2 x-3)^{2}=-12$.

$(2 x-3)^{2}=-12$

$2 x-3=\pm \sqrt{-12}$

$2 x-3=\pm 2 \sqrt{3} i$

Add $3$ to both sides.

$2 x=3 \pm 2 \sqrt{3} i$

$x=\frac{3 \pm 2 \sqrt{3 i}}{2}$

Rewrite in standard form.

$x=\frac{3}{2} \pm \frac{2 \sqrt{3} i}{2}$

$x=\frac{3}{2} \pm \sqrt{3} i$

$x=\frac{3}{2}+\sqrt{3} i, x=\frac{3}{2}-\sqrt{3} i$

Exercise $\PageIndex{17}$

Solve: $(3 r+4)^{2}=-8$.

$r=-\frac{4}{3}+\frac{2 \sqrt{2} i}{3}, r=-\frac{4}{3}-\frac{2 \sqrt{2} i}{3}$

Exercise $\PageIndex{18}$

Solve: $(2 t-8)^{2}=-10$.

$t=4+\frac{\sqrt{10} i}{2}, t=4-\frac{\sqrt{10 i}}{2}$

The left sides of the equations in the next two examples do not seem to be of the form $a(x-h)^{2}$. But they are perfect square trinomials, so we will factor to put them in the form we need.

Example $\PageIndex{10}$

Solve: $4 n^{2}+4 n+1=16$.

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

Exercise $\PageIndex{19}$

Solve: $9 m^{2}-12 m+4=25$.

$m=\frac{7}{3}, \quad m=-1$

Exercise $\PageIndex{20}$

Solve: $16 n^{2}+40 n+25=4$.

$n=-\frac{3}{4}, \quad n=-\frac{7}{4}$

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

Key Concepts

If $x^{2}=k$, then $x=\sqrt{k}$ or $x=-\sqrt{k}$or $x=\pm \sqrt{k}$

Learning Objectives

By the end of this section, you will be able to:

Be Prepared 9.1

Before you get started, take this readiness quiz.

Simplify: 128 . 128 . If you missed this problem, review Example 8.13 .

Be Prepared 9.2

Simplify: 32 5 32 5 . If you missed this problem, review Example 8.50 .

Be Prepared 9.3

Factor: 9 x 2 − 12 x + 4 9 x 2 − 12 x + 4 . If you missed this problem, review Example 6.23 .

A quadratic equation is an equation of the form ax 2 + bx + c = 0, where a ≠ 0 a ≠ 0 . Quadratic equations differ from linear equations by including a quadratic term with the variable raised to the second power of the form ax 2 . We use different methods to solve quadratic equations than linear equations, because just adding, subtracting, multiplying, and dividing terms will not isolate the variable.

We have seen that some quadratic equations can be solved by factoring. In this chapter, we will learn three other methods to use in case a quadratic equation cannot be factored.

Solve Quadratic Equations of the form a x 2 = k a x 2 = k using the Square Root Property

We have already solved some quadratic equations by factoring. Let’s review how we used factoring to solve the quadratic equation x 2 = 9.

We can easily use factoring to find the solutions of similar equations, like x 2 = 16 and x 2 = 25, because 16 and 25 are perfect squares. In each case, we would get two solutions, x = 4 , x = −4 x = 4 , x = −4 and x = 5 , x = −5 . x = 5 , x = −5 .

But what happens when we have an equation like x 2 = 7? Since 7 is not a perfect square, we cannot solve the equation by factoring.

Previously we learned that since 169 is the square of 13, we can also say that 13 is a square root of 169. Also, (−13) 2 = 169, so −13 is also a square root of 169. Therefore, both 13 and −13 are square roots of 169. So, every positive number has two square roots—one positive and one negative. We earlier defined the square root of a number in this way:

Since these equations are all of the form x 2 = k , the square root definition tells us the solutions are the two square roots of k . This leads to the Square Root Property .

Square Root Property

If x 2 = k , then

Notice that the Square Root Property gives two solutions to an equation of the form x 2 = k , the principal square root of k k and its opposite. We could also write the solution as x = ± k . x = ± k . We read this as x equals positive or negative the square root of k .

Now we will solve the equation x 2 = 9 again, this time using the Square Root Property.

What happens when the constant is not a perfect square? Let’s use the Square Root Property to solve the equation x 2 = 7.

We cannot simplify 7 7 , so we leave the answer as a radical.

Example 9.1

How to solve a quadratic equation of the form ax 2 = k using the square root property.

Solve: x 2 − 50 = 0 . x 2 − 50 = 0 .

Step one is to isolate the quadratic term and make its coefficient one. For the equation x squared minus fifty equals zero, first add fifty to both sides to get x squared by itself. The new equation is x squared equals fifty.

Solve: x 2 − 48 = 0 . x 2 − 48 = 0 .

Solve: y 2 − 27 = 0 . y 2 − 27 = 0 .

The steps to take to use the Square Root Property to solve a quadratic equation are listed here.

Solve a quadratic equation using the square root property.

In order to use the Square Root Property, the coefficient of the variable term must equal one. In the next example, we must divide both sides of the equation by the coefficient 3 before using the Square Root Property.

Example 9.2

Solve: 3 z 2 = 108 . 3 z 2 = 108 .

Solve: 2 x 2 = 98 . 2 x 2 = 98 .

Solve: 5 m 2 = 80 . 5 m 2 = 80 .

The Square Root Property states ‘If x 2 = k x 2 = k ,’ What will happen if k < 0 ? k < 0 ? This will be the case in the next example.

Example 9.3

Solve: x 2 + 72 = 0 x 2 + 72 = 0 .

Solve: c 2 + 12 = 0 . c 2 + 12 = 0 .

Solve: q 2 + 24 = 0 . q 2 + 24 = 0 .

Our method also works when fractions occur in the equation; we solve as any equation with fractions. In the next example, we first isolate the quadratic term, and then make the coefficient equal to one.

Example 9.4

Solve: 2 3 u 2 + 5 = 17 . 2 3 u 2 + 5 = 17 .

Solve: 1 2 x 2 + 4 = 24 . 1 2 x 2 + 4 = 24 .

Solve: 3 4 y 2 − 3 = 18 . 3 4 y 2 − 3 = 18 .

The solutions to some equations may have fractions inside the radicals. When this happens, we must rationalize the denominator .

Example 9.5

Solve: 2 x 2 − 8 = 41 . 2 x 2 − 8 = 41 .

Solve: 5 r 2 − 2 = 34 . 5 r 2 − 2 = 34 .

Try It 9.10

Solve: 3 t 2 + 6 = 70 . 3 t 2 + 6 = 70 .

Solve Quadratic Equations of the Form a ( x − h ) 2 = k Using the Square Root Property

We can use the Square Root Property to solve an equation of the form a ( x − h ) 2 = k as well. Notice that the quadratic term, x , in the original form ax 2 = k is replaced with ( x − h ).

On the left is the equation a times x square equals k. Replacing x in this equation with the expression x minus h changes the equation. It is now a times the square of x minus h equals k.

The first step, like before, is to isolate the term that has the variable squared. In this case, a binomial is being squared. Once the binomial is isolated, by dividing each side by the coefficient of a , then the Square Root Property can be used on ( x − h ) 2 .

Example 9.6

Solve: 4 ( y − 7 ) 2 = 48 . 4 ( y − 7 ) 2 = 48 .

Try It 9.11

Solve: 3 ( a − 3 ) 2 = 54 . 3 ( a − 3 ) 2 = 54 .

Try It 9.12

Solve: 2 ( b + 2 ) 2 = 80 . 2 ( b + 2 ) 2 = 80 .

Remember when we take the square root of a fraction, we can take the square root of the numerator and denominator separately.

Example 9.7

Solve: ( x − 1 3 ) 2 = 5 9 . ( x − 1 3 ) 2 = 5 9 .

Try It 9.13

Solve: ( x − 1 2 ) 2 = 5 4 . ( x − 1 2 ) 2 = 5 4 .

Try It 9.14

Solve: ( y + 3 4 ) 2 = 7 16 . ( y + 3 4 ) 2 = 7 16 .

We will start the solution to the next example by isolating the binomial term.

Example 9.8

Solve: 2 ( x − 2 ) 2 + 3 = 57 . 2 ( x − 2 ) 2 + 3 = 57 .

Try It 9.15

Solve: 5 ( a − 5 ) 2 + 4 = 104 . 5 ( a − 5 ) 2 + 4 = 104 .

Try It 9.16

Solve: 3 ( b + 3 ) 2 − 8 = 88 . 3 ( b + 3 ) 2 − 8 = 88 .

Sometimes the solutions are complex numbers.

Example 9.9

Solve: ( 2 x − 3 ) 2 = −12 . ( 2 x − 3 ) 2 = −12 .

Try It 9.17

Solve: ( 3 r + 4 ) 2 = −8 . ( 3 r + 4 ) 2 = −8 .

Try It 9.18

Solve: ( 2 t − 8 ) 2 = −10 . ( 2 t − 8 ) 2 = −10 .

The left sides of the equations in the next two examples do not seem to be of the form a ( x − h ) 2 . But they are perfect square trinomials, so we will factor to put them in the form we need.

Example 9.10

Solve: 4 n 2 + 4 n + 1 = 16 . 4 n 2 + 4 n + 1 = 16 .

We notice the left side of the equation is a perfect square trinomial. We will factor it first.

Try It 9.19

Solve: 9 m 2 − 12 m + 4 = 25 . 9 m 2 − 12 m + 4 = 25 .

Try It 9.20

Solve: 16 n 2 + 40 n + 25 = 4 . 16 n 2 + 40 n + 25 = 4 .

Access this online resource for additional instruction and practice with using the Square Root Property to solve quadratic equations.

Section 9.1 Exercises

Practice makes perfect.

Solve Quadratic Equations of the Form ax 2 = k Using the Square Root Property

In the following exercises, solve each equation.

a 2 = 49 a 2 = 49

b 2 = 144 b 2 = 144

r 2 − 24 = 0 r 2 − 24 = 0

t 2 − 75 = 0 t 2 − 75 = 0

u 2 − 300 = 0 u 2 − 300 = 0

v 2 − 80 = 0 v 2 − 80 = 0

4 m 2 = 36 4 m 2 = 36

3 n 2 = 48 3 n 2 = 48

4 3 x 2 = 48 4 3 x 2 = 48

5 3 y 2 = 60 5 3 y 2 = 60

x 2 + 25 = 0 x 2 + 25 = 0

y 2 + 64 = 0 y 2 + 64 = 0

x 2 + 63 = 0 x 2 + 63 = 0

y 2 + 45 = 0 y 2 + 45 = 0

4 3 x 2 + 2 = 110 4 3 x 2 + 2 = 110

2 3 y 2 − 8 = −2 2 3 y 2 − 8 = −2

2 5 a 2 + 3 = 11 2 5 a 2 + 3 = 11

3 2 b 2 − 7 = 41 3 2 b 2 − 7 = 41

7 p 2 + 10 = 26 7 p 2 + 10 = 26

2 q 2 + 5 = 30 2 q 2 + 5 = 30

5 y 2 − 7 = 25 5 y 2 − 7 = 25

3 x 2 − 8 = 46 3 x 2 − 8 = 46

( u − 6 ) 2 = 64 ( u − 6 ) 2 = 64

( v + 10 ) 2 = 121 ( v + 10 ) 2 = 121

( m − 6 ) 2 = 20 ( m − 6 ) 2 = 20

( n + 5 ) 2 = 32 ( n + 5 ) 2 = 32

( r − 1 2 ) 2 = 3 4 ( r − 1 2 ) 2 = 3 4

( x + 1 5 ) 2 = 7 25 ( x + 1 5 ) 2 = 7 25

( y + 2 3 ) 2 = 8 81 ( y + 2 3 ) 2 = 8 81

( t − 5 6 ) 2 = 11 25 ( t − 5 6 ) 2 = 11 25

( a − 7 ) 2 + 5 = 55 ( a − 7 ) 2 + 5 = 55

( b − 1 ) 2 − 9 = 39 ( b − 1 ) 2 − 9 = 39

4 ( x + 3 ) 2 − 5 = 27 4 ( x + 3 ) 2 − 5 = 27

5 ( x + 3 ) 2 − 7 = 68 5 ( x + 3 ) 2 − 7 = 68

( 5 c + 1 ) 2 = −27 ( 5 c + 1 ) 2 = −27

( 8 d − 6 ) 2 = −24 ( 8 d − 6 ) 2 = −24

( 4 x − 3 ) 2 + 11 = −17 ( 4 x − 3 ) 2 + 11 = −17

( 2 y + 1 ) 2 − 5 = −23 ( 2 y + 1 ) 2 − 5 = −23

m 2 − 4 m + 4 = 8 m 2 − 4 m + 4 = 8

n 2 + 8 n + 16 = 27 n 2 + 8 n + 16 = 27

x 2 − 6 x + 9 = 12 x 2 − 6 x + 9 = 12

y 2 + 12 y + 36 = 32 y 2 + 12 y + 36 = 32

25 x 2 − 30 x + 9 = 36 25 x 2 − 30 x + 9 = 36

9 y 2 + 12 y + 4 = 9 9 y 2 + 12 y + 4 = 9

36 x 2 − 24 x + 4 = 81 36 x 2 − 24 x + 4 = 81

64 x 2 + 144 x + 81 = 25 64 x 2 + 144 x + 81 = 25

Mixed Practice

In the following exercises, solve using the Square Root Property.

2 r 2 = 32 2 r 2 = 32

4 t 2 = 16 4 t 2 = 16

( a − 4 ) 2 = 28 ( a − 4 ) 2 = 28

( b + 7 ) 2 = 8 ( b + 7 ) 2 = 8

9 w 2 − 24 w + 16 = 1 9 w 2 − 24 w + 16 = 1

4 z 2 + 4 z + 1 = 49 4 z 2 + 4 z + 1 = 49

a 2 − 18 = 0 a 2 − 18 = 0

b 2 − 108 = 0 b 2 − 108 = 0

( p − 1 3 ) 2 = 7 9 ( p − 1 3 ) 2 = 7 9

( q − 3 5 ) 2 = 3 4 ( q − 3 5 ) 2 = 3 4

m 2 + 12 = 0 m 2 + 12 = 0

n 2 + 48 = 0 . n 2 + 48 = 0 .

u 2 − 14 u + 49 = 72 u 2 − 14 u + 49 = 72

v 2 + 18 v + 81 = 50 v 2 + 18 v + 81 = 50

( m − 4 ) 2 + 3 = 15 ( m − 4 ) 2 + 3 = 15

( n − 7 ) 2 − 8 = 64 ( n − 7 ) 2 − 8 = 64

( x + 5 ) 2 = 4 ( x + 5 ) 2 = 4

( y − 4 ) 2 = 64 ( y − 4 ) 2 = 64

6 c 2 + 4 = 29 6 c 2 + 4 = 29

2 d 2 − 4 = 77 2 d 2 − 4 = 77

( x − 6 ) 2 + 7 = 3 ( x − 6 ) 2 + 7 = 3

( y − 4 ) 2 + 10 = 9 ( y − 4 ) 2 + 10 = 9

Writing Exercises

In your own words, explain the Square Root Property.

In your own words, explain how to use the Square Root Property to solve the quadratic equation ( x + 2 ) 2 = 16 ( x + 2 ) 2 = 16 .

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table provides a checklist to evaluate mastery of the objectives of this section. Choose how would you respond to the statement “I can solve quadratic equations of the form a times x squared equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.” Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

Choose how would you respond to the statement “I can solve quadratic equations of the form a times the square of x minus h equals k using the Square Root Property.” “Confidently,” “with some help,” or “No, I don’t get it.”

ⓑ If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Kuta Software - Infinite Algebra 2. Name___________________________________. Period____. Date________________. Square Root Equations. Solve each equation.

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1. Solved example of equations with square roots. 3 t - 2 0 \sqrt{3t}-2=0 2. The power of a product is equal to the product of it's factors raised to the

9.6 Solve Equations with Square Roots

Solving radical equations.

Square Root Equation Calculator is a free online tool that displays the variable for the given square root equation. BYJU'S online square root equation

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How to Solve Radical Equations

Last Updated: March 11, 2023 References

This article was co-authored by wikiHow Staff . Our trained team of editors and researchers validate articles for accuracy and comprehensiveness. wikiHow's Content Management Team carefully monitors the work from our editorial staff to ensure that each article is backed by trusted research and meets our high quality standards. This article has been viewed 70,474 times. Learn more...

${\sqrt {x}}$

Solving Equations with One Radical

Image titled Solve Radical Equations Step 1

Final Answer: 64

${\sqrt[ {3}]{64}}-1=4-1=3$

The expression above was expanded through Polynomial Multiplication. If you're confused how it was done, you can review the process here. [4] X Research source

Solving Equations with Multiple Radicals

Image titled Solve Radical Equations Step 6

You frequently end up with quadratic equations when working with radicals. Review how to solve them if you are unsure.

Image titled Solve Radical Equations Step 7

Using the quadratic equation, you only get two possible answers: 2.53 and 11.47.

Image titled Solve Radical Equations Step 12

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Algebraically Find the Intersection of Two Lines

About This Article

To solve radical equations, which are any equations where the variable is under a square root, start by isolating the variable and radical on one side of the equation. Then, to undo the radical, square both sides of the equation. Check your answer by putting it back in the original equation. If the equation involves more complicated radicals, like the third root, follow the same procedure, but remove the radical by taking both sides of the equation to the third power. To learn how to use the same basic process for equations with multiple radicals, keep reading! Did this summary help you? Yes No

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How to Solve a Square Root Equation

How to Find the Line of Symmetry in a Quadratic Equation

The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 0 is 0, the square root of 100 is 10 and the square root of 50 is 7.071. Sometimes, you can figure out, or simply recall, the square root of a number that itself is a "perfect square," which is the product of an integer multiplied by itself; as you progress through your studies, you're likely to develop a mental list of these numbers (1, 4, 9, 25, 36 . . .).

Problems involving square roots are indispensable in engineering, calculus and virtually every realm of the modern world. Although you can easily locate square root equation calculators online (see Resources for an example), solving square root equations is an important skill in algebra, because it allows you to become familiar with using radicals and work with a number of problem types outside the realm of square roots per se.

Squares and Square Roots: Basic Properties

The fact that multiplying two negative numbers together yields a positive number is important in the world of square roots because it implies that positive numbers actually have two square roots (for example, the square roots of 16 are 4 and −4, even if only the former is intuitive). Similarly, negative numbers do not have real square roots, because there is no real number that takes on a negative value when multiplied by itself. In this presentation, the negative square root of a positive number will be ignored, so that "square root of 361" can be taken as "19" rather than " −19 and 19."

Also, when trying to estimate the value of a square root when no calculator is handy, it is important to realize that functions involving squares and square roots are not linear. You'll see more on this in the section about graphs later, but as a rough example, you have already observed that the square root of 100 is 10 and the square root of 0 is 0. On sight, this might lead you to guess that the square root for 50 (which is halfway between 0 and 100) must be 5 (which is halfway between 0 and 10). But you have also already learned that the square root of 50 is 7.071.

Finally, you may have internalized the idea that multiplying two numbers together yields a number greater than itself, implying that square roots of numbers are always smaller than the original number. This is not the case! Numbers between 0 and 1 have square roots, too, and in every case, the square root is greater than the original number. This is most easily shown using fractions. For example, 16/25, or 0.64, has a perfect square in both the numerator and the denominator. This means that the square root of the fraction is the square root of its top and bottom components, which is 4/5. This is equal to 0.80, a greater number than 0.64.

Square Root Terminology

"The square root of x " is usually written using what is called a radical sign, or just a radical (√ ). Thus for any x :

represents its square root. Flipping this around, the square of a number x is written using an exponent of 2 ( x 2 ). Exponents take superscripts on word-processing and related applications, and are also called powers. Because radical signs are not always easy to produce on demand, another way to write "the square root of x " is to use an exponent:

This in turn is part of a general scheme:

means "raise x to the power of y , then take the ' z ' root of it." x 1/2 thus means "raise x to the first power, which is simply x again, and then take the 2 root of it, or the square root." Extending this, x (5/3) means "raise x to the power of 5, then find the third root (or cube root) of the result."

Radicals can be used to represent roots other than 2, the square root. This is done by simply appending a superscript to the upper left of the radical.

then, represents the same number as x (5/3) from the previous paragraph does.

Most square roots are irrational numbers. This means that not only are they not nice, neat integers (e.g., 1, 2, 3, 4 . . .), but they also cannot be expressed as a neat decimal number that terminates without having to be rounded off. A rational number can be expressed as a fraction. So even though 2.75 is not an integer, it is a rational number because it is the same thing as the fraction 11/4. You were told earlier that the square root of 50 is 7.071, but this is actually rounded off from an infinite number of decimal places. The exact value of √50 is 5√2, and you'll see how this is determined soon.

Graphs of Square Root Functions

You have already seen that equations in involving squares and square roots are nonlinear. One easy way to remember this is that the graphs of the solutions of these equations are not lines. This makes sense, because if, as noted, the square of 0 is 0 and the square of 10 is 100 but the square of 5 is not 50, the graph resulting from simply squaring a number must curve its way to the correct values.

This is the case with the graph of

as you can see for yourself by visiting the calculator in the Resources and changing the parameters. The line passes through the point (0,0), and y does not go below 0, which you should expect because you know that x 2 is never negative. You can also see that the graph is symmetrical around the y -axis, which also makes sense because every positive square root of a given number is accompanied by a negative square root of equal magnitude. Therefore, with the exception of 0, every y value on the graph of y = x 2 is associated with two x -values.

Square Root Problems

One way to tackle basic square root problems by hand is to look for perfect squares "hidden" inside the problem. First, it's important to be aware of a few vital properties of squares and square roots. One of these is that, just as √ x 2 is simply equal to x (because the radical and the exponent cancel each other out):

That is, if you have a perfect square under a radical multiplying another number, you can "pull it out" and use it as a coefficient of what remains. For example, returning to the square root of 50

Sometimes you can wind up with a number involving square roots that is expressed as a fraction, but is still an irrational number because the denominator, the numerator or both contain a radical. In such instances, you may be asked to rationalize the denominator. For example, the number

has a radical in both the numerator and the denominator. But after scrutinizing "45," you may recognize it as the product of 9 and 5, which means that

Therefore, the fraction can be written

The radicals cancel each other out, and you are left with 6/3 = 2.

The basics of square roots (examples & answers), what are radicals in math, how to find the square root of an irrational number, definition of successor and predecessor in math, what is an inequality, how to divide negative numbers, what is the difference between integers and real numbers, how to write 5/6 as a mixed number or a decimal, when solving quadratic equations, what questions should..., what is function notation, how to find cube root in ti-84, how to solve cubic polynomials, how to find the roots of a quadratic, how to simplify radical fractions, how to calculate the slope of a tangent, how to get rid of a variable that is cubed, the difference between scientific & engineering notation, how to do integers on the calculator, how to get rid of a square root in an equation.

About the Author

Kevin Beck holds a bachelor's degree in physics with minors in math and chemistry from the University of Vermont. Formerly with ScienceBlogs.com and the editor of "Run Strong," he has written for Runner's World, Men's Fitness, Competitor, and a variety of other publications. More about Kevin and links to his professional work can be found at www.kemibe.com.

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Solving quadratic equations - Edexcel test questions - Edexcel

Solve 2x 2 = 162

Solve 3x 2 - 5 = 43

How do you factorise the following quadratic: x 2 - 5x - 14?

(x - 7) (x + 2)

(x - 2)(x + 7)

(x - 5)(x - 14)

Solve x 2 - 5x - 14 = 0

x = -5 and x = -14

x = -7 and x = 2

x = 7 and x = -2

Factorise the quadratic 2x 2 + x - 3

(2x - 2)(4x - 4)

(2x + 3)(x - 1)

(2x + 1)(x - 3)

Solve 2x 2 + x - 3 = 0

\[x = 1\:and\:x = \frac{-3}{2}\]

\[x = \frac{3}{2}\:and\:x = -1\]

\[x = 1\:and\:x = \frac{3}{2}\]

Write x 2 + 5x in completed square form.

(x + 5) 2 - 25

\[(x + \frac{5}{2})^2 - \frac{25}{4}\]

\[(x - 5)^2 + \frac{25}{4}\]

Solve x 2 + 10x + 6 = 0

\[x = 5 \pm \sqrt{19}\]

\[x = 19 \pm \sqrt{5}\]

\[x = -5 \pm \sqrt{19}\]

In the quadratic 2x 2 - 7x - 7 = 0, what are the values of a, b and c?

a = 2, b = -7, c = -7

a = 2, b = 7 and c = 7

a = 0, b = 7 and c = 7

Solve 2x 2 - 7x - 7 = 0, leaving the answer in surd (square root) form.

\[x = \frac{7 \pm \sqrt{105}}{4}\]

\[x = \frac{7 \pm \sqrt{105}}{2}\]

\[x = \frac{-7 \pm \sqrt{105}}{4}\]

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IMAGES

9.1.1 Solving Quadratic Equations Using the Square Root Property
Solving a Square Root Equation Example 3 (Two Radicals)
Solving an equation that contains a square root
Algebra: Equations with Square Roots
solve quadratic by square roots
👍 Solving square root problems. How to Solve Square Root Problems (with Pictures). 2019-02-19

VIDEO

10% of Math Students can solve this equation
10.1 Video Lecture Notes
Solving Square Root Equations
Solve the equations: Square root(x)- Square root(y)=9 and x-y=9,value of x and y ? @mantubera9073
Solving Square Root Equations with and without Extraneous Solutions
Square Roots Equations #maths #algebra

COMMENTS

Solving square-root equations (article)
We can use a linear approximation to find a close estimate for the square root. √ (x) ≈ (x + y) / (2 * √ (y)) where y is a number that is "close to" x. Typically, you would choose y to be a perfect square to make the math easy. ( 3 votes) svmejia 7 years ago
Solving square-root equations: one solution
If you mean x^2 = 8, then you have to transpose the square; look x^2 = 8 x = sqrt (8) x = [sqrt (4)] [sqrt (2)] // Using the product rule [ sqrt (a)*sqrt (b) = sqrt (ab)] look for perfect squares and simplify x = 2 [sqrt (2)] // sqrt (4) = 2 since 2^2 =4
8.6: Solve Equations with Square Roots
Now we will see how to solve a radical equation. Our strategy is based on the relation between taking a square root and squaring. For a ≥ 0, (√a)2 = a How to Solve Radical Equations Example 8.6.4 Solve: √2x − 1 = 7 Answer Example 8.6.5 Solve: √3x − 5 = 5. Answer Example 8.6.6 Solve: √4x + 8 = 6. Answer Definition: SOLVE A RADICAL EQUATION.
How to Solve Square Root Problems (with Pictures)
Now, since 9, which is a perfect square, is separated from 100, we can take its square root on its own. √ (9 × 100) = √ (9) × √ (100) = 3 × √ (100). In other words, √ (900) = 3√ (100). We can even simplify this two steps further by dividing 100 into the factors 25 and 4. √ (100) = √ (25 × 4) = √ (25) × √ (4) = 5 × 2 = 10.
10.1 Solve Quadratic Equations Using the Square Root Property
Introduction; 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality; 2.2 Solve Equations using the Division and Multiplication Properties of Equality; 2.3 Solve Equations with Variables and Constants on Both Sides; 2.4 Use a General Strategy to Solve Linear Equations; 2.5 Solve Equations with Fractions or Decimals; 2.6 Solve a Formula for a Specific Variable
Solving Quadratic Equations by Square Root Method
Key Strategy in Solving Quadratic Equations using the Square Root Method The general approach is to collect all {x^2} x2 terms on one side of the equation while keeping the constants to the opposite side. After doing so, the next obvious step is to take the square roots of both sides to solve for the value of x x.
Square-root equations (practice)
Learn for free about math, art, computer programming, economics, physics, chemistry, biology, medicine, finance, history, and more. Khan Academy is a nonprofit with the mission of providing a free, world-class education for anyone, anywhere.
Square Roots Calculator
Square Roots Calculator Find square roots of any number step-by-step full pad » Examples Related Symbolab blog posts Practice, practice, practice Math can be an intimidating subject. Each new topic we learn has symbols and problems we have never seen. The unknowing... Read More
Solve Equations with Roots
The strategy for solving is to isolate the square root on the left side of the equation and then square both sides. First subtract 2 from both sides: x − 3 = 4. Now that the square root is isolated, we can square both sides of the equation: ( x − 3) 2 = 4 2. Since the square and the square root cancel we get: x − 3 = 16.
9.2: Solve Quadratic Equations Using the Square Root Property
Solve a Quadratic Equation Using the Square Root Property Isolate the quadratic term and make its coefficient one. Use Square Root Property. Simplify the radical. Check the solutions. In order to use the Square Root Property, the coefficient of the variable term must equal one.
9.1 Solve Quadratic Equations Using the Square Root Property
Solve Quadratic Equations of the Form a(x − h) 2 = k Using the Square Root Property. We can use the Square Root Property to solve an equation of the form a(x − h) 2 = k as well. Notice that the quadratic term, x, in the original form ax 2 = k is replaced with (x − h). The first step, like before, is to isolate the term that has the variable squared.
Solving Quadratic Equations Using Square Roots
This video explains how use square roots to solve quadratic equations in the form ax^2+c=0.
Solve Quadratic By Square Root Teaching Resources
4.8. (33) $4.50. PDF. Students will solve 14 quadratic equations (where b=0) using square roots. There are three levels included to provide easy differentiation for your classroom (solutions as approximate values, solutions as exact values and solutions as exact values plus four multi-step equations). Two unique formats, printable coloring ...
Ex 4: Solve Radical Equations
98K views 11 years ago Solving Radical Equations This video provides two examples of how to solve a radical equations containing square roots with the variable under the square root...
Solve square root equations
Solve square root equations - We can use a linear approximation to find a close estimate for the square root. (x) (x + y) / (2 * (y)) where y is a number that. ... Square Root Equation Calculator is a free online tool that displays the variable for the given square root equation. BYJU'S online square root equation
How to Solve Radical Equations
2. Square both sides of the equation to remove the radical. All you have to do to undo a radical is square it. Because you need the equation to stay balanced, you square both sides, just like you added or subtracted from both sides earlier. So, for the example: Isolate. x {\displaystyle {\sqrt {x}}}
Solving Quadratics Quiz 2
Solving Quadratics Quiz 2. Q. Solve the equation using square roots. Solve the equation using square roots. Use the square root property to solve the Quadratic Equation. Q. Use the square root property to solve the Quadratic Equation. Solve the equation using the zero product property.
Algebra Calculator
Square Root In mathematics, a square root of a number x is a number y such that y² = x; in other words, a number y whose square (the result of multiplying the number by itself, or y ⋅ y) is x. For example, 4 and −4 are square roots of 16, because 4² = (−4)² = 16.
How to Solve a Square Root Equation
This in turn is part of a general scheme: x^ { (y/z)} x(y/z) means "raise x to the power of y , then take the ' z ' root of it." x 1/2 thus means "raise x to the first power, which is simply x again, and then take the 2 root of it, or the square root." Extending this, x (5/3) means "raise x to the power of 5, then find the third root (or cube ...
Solving quadratic equations
Solving quadratic equations. Solve quadratic equations by factorising, using formulae and completing the square. Each method also provides information about the corresponding quadratic graph. Part of.