Square Roots and Completing the Square
Learning Outcomes
- Use the square root property to solve a quadratic equation
- Complete the square to solve a quadratic equation
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 that would let you solve this 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.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. Do not forget to use both positive and negative square roots!
[latex]\begin{array}{c}x^{2}=9 \\ x=\pm\sqrt{9} \end{array}[/latex]
[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])
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 cannot simplify that. Subtract [latex]5[/latex] from both sides to get the [latex]x^{2}[/latex] term by itself.
[latex]10x^{2}=80[/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]x^{2}=8[/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}{ll}{{x}^{2}} & =8\\ x & =\pm \sqrt{8}\\ & =\pm \sqrt{(4)(2)}\\ & =\pm \sqrt{4}\sqrt{2}\\ & =\pm 2\sqrt{2}\end{array}[/latex]
The answer is [latex] x=\pm 2\sqrt{2}[/latex].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 cannot work with on the left. Start by adding 50 to both sides.
[latex]\left(x-2\right)^{2}=50[/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 [latex]x[/latex] on the left, you need to add [latex]2[/latex] to both sides. Be sure to simplify the radical if possible.[latex] \begin{array}{ll}x & =2\pm \sqrt{50} \\ & =2\pm \sqrt{(25)(2)} \\ & =2\pm \sqrt{25}\sqrt{2} \\ & =2\pm 5\sqrt{2}\end{array}[/latex]
The answer is [latex] x=2\pm 5\sqrt{2}[/latex].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 us make sure we can recognize a perfect square trinomial and how to 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 these squared 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]
Steps for Completing The Square
We will use the example [latex]{x}^{2}+4x+1=0[/latex] to illustrate each step.- 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]
- Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
[latex]\begin{array}{c}\frac{1}{2}\left(4\right)=2\hfill \\ {2}^{2}=4\hfill \end{array}[/latex]
- 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]
- The left side of the equation can now be factored as a perfect square.
[latex]\begin{array}{c}{x}^{2}+4x+4=3\hfill \\ {\left(x+2\right)}^{2}=3\hfill \end{array}[/latex]
- Use the square root property and solve.
[latex]\begin{array}{c}\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]
- 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.
[latex]\begin{array}{r}x^{2}-12x=4\end{array}[/latex]
Identify [latex]b[/latex]: [latex]b=-12[/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 will miss one of the solutions.[latex] x-6=\pm\sqrt{40}[/latex]
Solve for [latex]x[/latex] 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]
The answer is [latex] x=6\pm 2\sqrt{10}[/latex].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.
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]
[latex]x=-8[/latex]
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 [latex]x[/latex]. 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.