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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 x2=9x^{2}=9 is to subtract 99 from both sides to get one side equal to 0: x29=0x^{2}-9=0. The expression on the left can be factored, it is a difference of squares: (x+3)(x3)=0\left(x+3\right)\left(x–3\right)=0. Using the zero factor property, you know this means x+3=0x+3=0 or x3=0x–3=0, so x=3x=−3 or 33. Another property would let you solve that equation more easily is called the square root property.

The Square Root Property

If x2=ax^{2}=a, then x=a x=\sqrt{a} or a -\sqrt{a}. 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 “a \sqrt{a}” or “a -\sqrt{a}” is ±a \pm \sqrt{a}. The symbol ±\pm 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. x2=9x^{2}=9

Answer: Since one side is simply x2x^{2}, 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!

x2=9   x=±9   x=±3\begin{array}{l}x^{2}=9\\\,\,\,x=\pm\sqrt{9}\\\,\,\,x=\pm3\end{array}

Answer

x=±3x=\pm3 (that is, x=3x=3 or 3-3)

Notice that there is a difference here in solving x2=9x^{2}=9 and finding 9 \sqrt{9}. For x2=9x^{2}=9, you are looking for all numbers whose square is 99. For 9 \sqrt{9}, you only want the principal (nonnegative) square root. The negative of the principal square root is 9 -\sqrt{9}; both would be ±9 \pm \sqrt{9}. 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 x2x^{2}. 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. 10x2+5=8510x^{2}+5=85

Answer: If you try taking the square root of both sides of the original equation, you will have 10x2+5 \sqrt{10{{x}^{2}}+5} on the left, and you can’t simplify that. Subtract 55 from both sides to get the x2x^{2} term by itself.

10x2+5=8510x^{2}+5=85

You could now take the square root of both sides, but you would have 10 \sqrt{10} as a coefficient, and you would need to divide by that coefficient. Dividing by 1010 before you take the square root will be a little easier.

10x2=8010x^{2}=80

Now you have only x2x^{2} on the left, so you can use the Square Root Property easily. Be sure to simplify the radical if possible.

x2=8   x=±8      =±(4)(2)      =±42      =±22 \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}

Answer

x=±22 x=\pm 2\sqrt{2}

Sometimes more than just the x is being squared:

Example

Solve. (x2)250=0\left(x–2\right)^{2}–50=0

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.

(x2)250=0\left(x-2\right)^{2}-50=0

Because (x2)2\left(x–2\right)^{2} is a squared quantity, you can take the square root of both sides.

(x2)2=50          x2=±50\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\\x-2=\pm\sqrt{50}\end{array}

To isolate x on the left, you need to add 22 to both sides. Be sure to simplify the radical if possible.

x=2±50    =2±(25)(2)    =2±252    =2±52 \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}

Answer

x=2±52 x=2\pm 5\sqrt{2}

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 9x224x+169x^{2}–24x+16.

Answer: First notice that the x2x^{2} term and the constant term are both perfect squares. 9x2=(3x)2   16=42\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\\,\,\,16=4^{2}\end{array} Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms. 24x=2(3x)(4)24x=2\left(3x\right)\left(4\right) A trinomial in the form r22rs+s2r^{2}-2rs+s^{2} can be factored as (rs)2(r–s)^{2}. In this case, the middle term is subtracted, so subtract r and s and square it to get (rs)2(r–s)^{2}.    r=3xs=49x224x+16=(3x4)2\begin{array}{c}\,\,\,r=3x\\s=4\\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}

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 x2+4x+1=0{x}^{2}+4x+1=0 to illustrate each step.
  1. Given a quadratic equation that cannot be factored, and with a=1a=1, first add or subtract the constant term to the right side of the equal sign.
    x2+4x=1{x}^{2}+4x=-1
  2. Multiply the b term by 12\frac{1}{2} and square it.
    12(4)=222=4\begin{array}{l}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}
  3. Add (12b)2{\left(\frac{1}{2}b\right)}^{2} to both sides of the equal sign and simplify the right side. We have
    x2+4x+4=1+4x2+4x+4=3\begin{array}{l}{x}^{2}+4x+4=-1+4\hfill \\ {x}^{2}+4x+4=3\hfill \end{array}
  4. The left side of the equation can now be factored as a perfect square.
    x2+4x+4=3(x+2)2=3\begin{array}{l}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}
  5. Use the square root property and solve.
    (x+2)2=±3x+2=±3x=2±3\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}
  6. The solutions are x=2+3x=-2+\sqrt{3}, x=23x=-2-\sqrt{3}.
 

Example

Solve by completing the square. x212x4=0x^{2}–12x–4=0

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.

x212x=4        b=12\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}

Then, take 12\frac{1}{2} of the b term and square it. Add (b2)2 {{\left( \frac{b}{2}\right)}^{2}} to complete the square, so (b2)2=(122)2=(6)2=36 {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36. Add the value to both sides of the equation and simplify.

x212x+36=4+36x212x+36=40\begin{array}{l}x^{2}-12x+36=4+36\\x^{2}-12x+36=40\end{array}

Rewrite the left side as a squared binomial.

(x6)2=40\left(x-6\right)^{2}=40

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

x6=±40 x-6=\pm\sqrt{40}

Solve for x by adding 66 to both sides. Simplify as needed.

x=6±40    =6±410    =6±210 \begin{array}{l}x=6\pm \sqrt{40}\\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\\,\,\,\,=6\pm 2\sqrt{10}\end{array}

Answer

x=6±210 x=6\pm 2\sqrt{10}

Example

Solve by completing the square: x23x5=0{x}^{2}-3x - 5=0.

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

x23x=5{x}^{2}-3x=5
Identify bb.   b=3b=-3 Then, take 12\frac{1}{2} of the b term and square it.
12(3)=32(32)2=94\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}
Add the result to both sides of the equal sign.
 x23x+(32)2=5+(32)2x23x+94=5+94\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}
Factor the left side as a perfect square and simplify the right side.
(x32)2=294{\left(x-\frac{3}{2}\right)}^{2}=\frac{29}{4}
Use the square root property and solve.
(x32)2=±294(x32)=±292x=32±292\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}
The solutions are x=3+292x=\frac{3+\sqrt{29}}{2}, x=3292x=\frac{3-\sqrt{29}}{2}.

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. x2+16x+17=47x^{2}+16x+17=-47.

Answer: Rewrite the equation so the left side has the form x2+bxx^{2}+bx. Identify b.

x2+16x=64b=16\begin{array}{c}x^{2}+16x=-64\\b=16\end{array}

Add (b2)2 {{\left( \frac{b}{2} \right)}^{2}}, which is (162)2=82=64 {{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64, to both sides.

x2+16x+64=64+64x2+16x+64=0\begin{array}{l}x^{2}+16x+64=-64+64\\x^{2}+16x+64=0\end{array}

Write the left side as a squared binomial.

(x+8)2=0\left(x+8\right)^{2}=0

Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 00 has only one root.

x+8=0x+8=0

Solve for x.

x=8x=-8

Answer

x=8x=-8

Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 4747 to both sides in the equation. The equation x2+16x+17=47x^{2}+16x+17=−47 becomes x2+16x+64=0x^{2}+16x+64=0. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 1616). It can be factored as (x+8)(x+8)=0(x+8)(x+8)=0, 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 x2+bxx^{2}+bx into a perfect square trinomial x2+bx+(b2)2 {{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}, which can be factored to (x+b2)2 {{\left( x+\frac{b}{2} \right)}^{2}}. When solving quadratic equations by completing the square, be careful to add (b2)2 {{\left( \frac{b}{2} \right)}^{2}} 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.

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