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 with irrational roots
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 solutions of [latex]x^2=a[/latex] are called the square roots of a.- When a is positive, [latex]a > 0[/latex], [latex]x^2=a[/latex] has two solutions, [latex]+\sqrt{a},-\sqrt{a}[/latex]. [latex]+\sqrt{a}[/latex] is the nonnegative square root of a, and [latex]-\sqrt{a}[/latex] is the negative square root of a.
- When a is negative, [latex]a < 0[/latex], [latex]x^2=a[/latex] has no solutions.
- When a is zero, [latex]a = 0[/latex], [latex]x^2=a[/latex] has one solution: [latex]a = 0[/latex]
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].Try It
[ohm_question]197330[/ohm_question]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. Some of the above examples have squared binomials: [latex]\left(1+r\right)^{2}[/latex] and [latex]\left(x–2\right)^{2}[/latex] are squared binomials. If you expand these, you get a perfect square trinomial. Perfect square trinomials have the form [latex]x^{2}+2xs+s^{2}[/latex] and can be factored as [latex]\left(x+s\right)^{2}[/latex], or they have the form [latex]x^{2}–2xs+s^{2}[/latex] and can be factored as [latex]\left(x–s\right)^{2}[/latex]. Let’s factor a perfect square trinomial into a squared binomial.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]
Example
Solve. [latex]4x^{2}+20x+25=8[/latex]Answer: Since there’s an x term, you can’t use the Square Root Property immediately (or even after adding or dividing by a constant). Notice, however, that the [latex]x^{2}[/latex] and constant terms on the left are both perfect squares: [latex]\left(2x\right)^{2}[/latex] and [latex]5^{2}[/latex]. Check the middle term: is it [latex]2\left(2x\right)\left(5\right)[/latex]? Yes!
[latex]4x^{2}+20x+25=8[/latex]
A trinomial in the form [latex]r^{2}+2rs+s^{2}[/latex] can be factored as [latex]\left(r+s\right)^{2}[/latex], so rewrite the left side as a squared binomial.[latex](2x+5)^{2}=8[/latex]
Now you can use the Square Root Property. Some additional steps are needed to isolate x.[latex] \begin{array}{r}2x+5=\pm \sqrt{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\\\2x=-5\pm \sqrt{8}\,\,\,\,\,\\\\x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{8}\end{array}[/latex]
Simplify the radical when possible.[latex] \begin{array}{l}x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{4}\sqrt{2}\\\\x=-\frac{5}{2}\pm \frac{1}{2}(2)\sqrt{2}\\\\x=-\frac{5}{2}\pm \sqrt{2}\end{array}[/latex]
Answer
[latex-display] x=-\frac{5}{2}\pm \sqrt{2}[/latex-display]Finding a Value that will Complete the Square in an Expression
To complete the square for an expression of the form [latex]x^{2}+bx[/latex]:- Identify the value of b;
- Calculate and add [latex]\left(\frac{b}{2}\right)^{2}[/latex].
Example
Find the number to add to [latex]x^{2}+8x[/latex] to make it a perfect square trinomial.Answer: First identify b if this has the form [latex]x^{2}+bx[/latex].
[latex]\begin{array}{c}x^{2}+8x\\b=8\end{array}[/latex]
To complete the square, add [latex]\left(\frac{b}{2}\right)^{2}[/latex].[latex]b=8[/latex], so [latex]\left(\frac{b}{2}\right)^{2}=\left(\frac{8}{2}\right)^{2}[/latex]
Simplify.[latex]\begin{array}{c}x^{2}+8x+\left(4\right)^{2}\\x^{2}+8x+16\end{array}[/latex]
Check that the result is a perfect square trinomial. [latex]\left(x+4\right)^{2}=x^{2}+4x+4x+16=x^{2}+8x+16[/latex], so it is.Answer
Adding [latex]+16[/latex] will make [latex]x^{2}+8x[/latex] a perfect square trinomial.Example
Rewrite [latex]x^{2}+6x=8[/latex] so that the left side is a perfect square trinomial.Answer: This equation has a constant of 8. Ignore it for now and focus on the [latex]x^{2}[/latex] and x terms on the left side of the equation. The left side has the form [latex]x^{2}+bx[/latex], so you can identify b.
[latex]\begin{array}{r}x^{2}+6x=8\\b=6\end{array}[/latex]
To complete the square, add [latex] {{\left( \frac{b}{2} \right)}^{2}}[/latex] to the left side. [latex-display]b=6[/latex], so [latex] {{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{6}{2} \right)}^{2}}={{3}^{2}}=9.[/latex-display] This is an equation, though, so you must add the same number to the right side as well.[latex]x^{2}+6x+9=8+9[/latex]
Simplify. Check that the left side is a perfect square trinomial. [latex]\begin{array}{r}\left(x+3\right)^{2}=x^{2}+3x+3x+9=x^{2}+6x+9\end{array}{r}[/latex], so it is.[latex]\begin{array}{r}x^{2}+6x+9=17\\x^{2}+6x+9=17\\(x+3)^{2}=17\end{array}[/latex]
Answer
[latex-display]x^{2}+6x+9=17[/latex-display]Example
Solve. [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. Rewrite the equation with the left side in the form [latex]x^{2}+bx[/latex], to prepare to complete the square. Identify b.
[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\\b=-12\end{array}[/latex]
Figure out what value to add to complete the square. 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 6 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]Steps for Completing The Square
- Given a quadratic equation with [latex]a=1[/latex], first add or subtract the constant term c to the right side of the equal sign.
[latex]{x}^{2}+bx=-c[/latex]
- Multiply the b term by [latex]\frac{1}{2}[/latex] and square it.
[latex]{\left(\frac{1}{2}b\right)}^{2}[/latex]
- Add [latex]{\left(\frac{1}{2}b\right)}^{2}[/latex] to both sides of the equal sign and simplify the right side
- The left side of the equation can now be factored as a perfect square. [latex]x^{2}+bx+\left(\frac{b}{2}\right)^{2}=\left(x+\frac{b}{2}\right)^{2}[/latex]
- Use the square root property to solve.
Try It
[ohm_question]147603[/ohm_question]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.Contribute!
Licenses & Attributions
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- 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.