We've updated our
Privacy Policy effective December 15. Please read our updated Privacy Policy and tap

TEXT
Unlock Solution Steps
Sign in to
Symbolab
Get full access to all Solution Steps for any math problem
By continuing, you agree to our Terms of Use
and have read our Privacy Policy

Study Guides > Intermediate Algebra

Read: Square Roots and Completing the Square

Learning Objectives

  • Use the square root property to solve a quadratic equation
  • Complete the square to solve a quadratic equation
Quadratic equations can be solved in many ways. You may already be familiar with factoring to solve some quadratic equations. However, not all quadratic equations can be factored. In this topic, you will use square roots to learn another way to solve quadratic equations—and this method will work with all quadratic equations.

Solve a Quadratic Equation by the Square Root Property

One way to solve the quadratic equation [latex]x^{2}=9[/latex] is to subtract [latex]9[/latex] from both sides to get one side equal to 0: [latex]x^{2}-9=0[/latex]. The expression on the left can be factored, it is a difference of squares: [latex]\left(x+3\right)\left(x–3\right)=0[/latex]. Using the zero factor property, you know this means [latex]x+3=0[/latex] or [latex]x–3=0[/latex], so [latex]x=−3[/latex] or [latex]3[/latex]. Another property would let you solve that equation more easily is called the square root property.

The Square Root Property

If [latex]x^{2}=a[/latex], then [latex] x=\sqrt{a}[/latex] or [latex] -\sqrt{a}[/latex]. The property above says that you can take the square root of both sides of an equation, but you have to think about two cases: the positive square root of a and the negative square root of a.
A shortcut way to write “[latex] \sqrt{a}[/latex]” or “[latex] -\sqrt{a}[/latex]” is [latex] \pm \sqrt{a}[/latex]. The symbol [latex]\pm[/latex] is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”

Example

Solve using the Square Root Property. [latex]x^{2}=9[/latex]

Answer: Since one side is simply [latex]x^{2}[/latex], you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!

[latex]\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}[/latex]

Answer

[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])

Notice that there is a difference here in solving [latex]x^{2}=9[/latex] and finding [latex] \sqrt{9}[/latex]. For [latex]x^{2}=9[/latex], you are looking for all numbers whose square is [latex]9[/latex]. For [latex] \sqrt{9}[/latex], you only want the principal (nonnegative) square root. The negative of the principal square root is [latex] -\sqrt{9}[/latex]; both would be [latex] \pm \sqrt{9}[/latex]. Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted! In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating [latex]x^{2}[/latex]. In our first video we will show more examples of using the square root property to solve a quadratic equation. https://youtu.be/Fj-BP7uaWrI

Example

Solve. [latex]10x^{2}+5=85[/latex]

Answer: If you try taking the square root of both sides of the original equation, you will have [latex] \sqrt{10{{x}^{2}}+5}[/latex] on the left, and you can’t simplify that. Subtract [latex]5[/latex] from both sides to get the [latex]x^{2}[/latex] term by itself.

[latex]10x^{2}+5=85[/latex]

You could now take the square root of both sides, but you would have [latex] \sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by [latex]10[/latex] before you take the square root will be a little easier.

[latex]10x^{2}=80[/latex]

Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily. Be sure to simplify the radical if possible.

[latex] \begin{array}{l}{{x}^{2}}=8\\\,\,\,x=\pm \sqrt{8}\\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]

Answer

[latex-display] x=\pm 2\sqrt{2}[/latex-display]

Sometimes more than just the x is being squared:

Example

Solve. [latex]\left(x–2\right)^{2}–50=0[/latex]

Answer: Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.

[latex]\left(x-2\right)^{2}-50=0[/latex]

Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.

[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}[/latex]

To isolate x on the left, you need to add [latex]2[/latex] to both sides. Be sure to simplify the radical if possible.

[latex] \begin{array}{l}x=2\pm \sqrt{50}\\\,\,\,\,=2\pm \sqrt{(25)(2)}\\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]

Answer

[latex-display] x=2\pm 5\sqrt{2}[/latex-display]

In the next video you will see more examples of using square roots to solve quadratic equations. https://youtu.be/4H5qZ_-8YM4

Solve a Quadratic Equation by Completing the Square

Not all quadratic equations can be factored or solved in their original form using the square root property. In these cases, we may use a method for solving a quadratic equation known as completing the square. Using this method, we add or subtract terms to both sides of the equation until we have a perfect square trinomial on one side of the equal sign. We then apply the square root property. First, let's make sure we can recognize a perfect square trinomial and how factor it.

Example

Factor [latex]9x^{2}–24x+16[/latex].

Answer: First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares. [latex-display]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array}[/latex-display] Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms. [latex-display]24x=2\left(3x\right)\left(4\right)[/latex-display] A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as [latex](r–s)^{2}[/latex]. In this case, the middle term is subtracted, so subtract r and s and square it to get [latex](r–s)^{2}[/latex]. [latex-display]\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex-display]

If this were an equation, we could solve using either the square root property or the zero product property. If you don't start with a perfect square trinomial, you can complete the square to make what you have into one. To complete the square, the leading coefficient, a, must equal 1. If it does not, then divide the entire equation by a. Then, we can use the following procedures to solve a quadratic equation by completing the square.

Steps for Completing The Square

We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.
  1. Given a quadratic equation that cannot be factored, and with [latex]a=1[/latex], first add or subtract the constant term to the right side of the equal sign.
    [latex]{x}^{2}+4x=-1[/latex]
  2. Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
    [latex]\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
  3. Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side. We have
    [latex]\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}[/latex]
  4. The left side of the equation can now be factored as a perfect square.
    [latex]\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
  5. Use the square root property and solve.
    [latex]\begin{array}{l}\sqrt{{\left(x+2\right)}^{2}}=\pm \sqrt{3}\hfill \\ x+2=\pm \sqrt{3}\hfill \\ x=-2\pm \sqrt{3}\hfill \end{array}[/latex]
  6. The solutions are [latex]x=-2+\sqrt{3}[/latex], [latex]x=-2-\sqrt{3}[/latex].
 

Example

Solve by completing the square. [latex]x^{2}–12x–4=0[/latex]

Answer: Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation. First, move the constant term to the right side of the equal sign.Identify b.

[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]

Then, take [latex]\frac{1}{2}[/latex] of the b term and square it. Add [latex] {{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex] {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex]. Add the value to both sides of the equation and simplify.

[latex]\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}[/latex]

Rewrite the left side as a squared binomial.

[latex]\left(x-6\right)^{2}=40[/latex]

Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.

[latex] x-6=\pm\sqrt{40}[/latex]

Solve for x by adding [latex]6[/latex] to both sides. Simplify as needed.

[latex] \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]

Answer

[latex-display] x=6\pm 2\sqrt{10}[/latex-display]

Example

Solve by completing the square: [latex]{x}^{2}-3x - 5=0[/latex].

Answer: First, move the constant term to the right side of the equal sign.

[latex]{x}^{2}-3x=5[/latex]
Identify [latex]b[/latex].   [latex]b=-3[/latex] Then, take [latex]\frac{1}{2}[/latex] of the b term and square it.
[latex]\begin{array}{l}\frac{1}{2}\left(-3\right)=-\frac{3}{2}\hfill \\ {\left(-\frac{3}{2}\right)}^{2}=\frac{9}{4}\hfill \end{array}[/latex]
Add the result to both sides of the equal sign.
[latex]\begin{array}{l}\text{ }{x}^{2}-3x+{\left(-\frac{3}{2}\right)}^{2}=5+{\left(-\frac{3}{2}\right)}^{2}\hfill \\ {x}^{2}-3x+\frac{9}{4}=5+\frac{9}{4}\hfill \end{array}[/latex]
Factor the left side as a perfect square and simplify the right side.
[latex]{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}[/latex]
Use the square root property and solve.
[latex]\begin{array}{l}\sqrt{{\left(x-\frac{3}{2}\right)}^{2}}\hfill&=\pm \sqrt{\frac{29}{4}}\hfill \\ \left(x-\frac{3}{2}\right)\hfill&=\pm \frac{\sqrt{29}}{2}\hfill \\ x\hfill&=\frac{3}{2}\pm \frac{\sqrt{29}}{2}\hfill \end{array}[/latex]
The solutions are [latex]x=\frac{3+\sqrt{29}}{2}[/latex], [latex]x=\frac{3-\sqrt{29}}{2}[/latex].

In the next video you will see more examples of how to use completing the square to solve a quadratic equation. https://youtu.be/PsbYUySRjFo You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.

Example

Solve by completing the square. [latex]x^{2}+16x+17=-47[/latex].

Answer: Rewrite the equation so the left side has the form [latex]x^{2}+bx[/latex]. Identify b.

[latex]\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}[/latex]

Add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex] {{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.

[latex]\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}[/latex]

Write the left side as a squared binomial.

[latex]\left(x+8\right)^{2}=0[/latex]

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. [latex]0[/latex] has only one root.

[latex]x+8=0[/latex]

Solve for x.

[latex]x=-8[/latex]

Answer

[latex-display]x=-8[/latex-display]

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding [latex]47[/latex] to both sides in the equation. The equation [latex]x^{2}+16x+17=−47[/latex] becomes [latex]x^{2}+16x+64=0[/latex]. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is [latex]16[/latex]). It can be factored as [latex](x+8)(x+8)=0[/latex], of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation. In our last video, we show an example of how to use completing the square to solve a quadratic equation whose solutions are rational. https://youtu.be/IjCjbtrPWHM

Summary

Completing the square is used to change a binomial of the form [latex]x^{2}+bx[/latex] into a perfect square trinomial [latex] {{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}[/latex], which can be factored to [latex] {{\left( x+\frac{b}{2} \right)}^{2}}[/latex]. When solving quadratic equations by completing the square, be careful to add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex] to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.

Licenses & Attributions

CC licensed content, Shared previously

  • College Algebra. Provided by: OpenStax Authored by: Abramson, Jay, et al.. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution. License terms: Download for free at: http://cnx.org/contents/[email protected]:1/Preface.
  • Ex 1: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex 2: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex 1: Completing the Square - Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.
  • Ex 2: Completing the Square - Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) for Lumen Learning. License: CC BY: Attribution.