Problems

Solve by completing the square

\(\textbf{1)}\) \( x^2+10x-24=0 \)

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The answer is \( x=-12,2 \)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2+10x-24=0\)
\(\,\,\,\,\,\,x^2+10x=24\)
\(\,\,\,\,\,\,x^2+10x+\text{___}=24+ \text{___}\)

\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{10}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(5\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle25\)
\(\,\,\,\,\,\,x^2+10x+\underline{25}=24+ \underline{25}\)
\(\,\,\,\,\,\,x^2+10x+25=49\)

\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x+5)^2=49\)

\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x+5)^2}=\sqrt{49}\)
\(\,\,\,\,\,\,x+5=\pm7\)
\(\,\,\,\,\,\,x=\pm7-5\)
\(\,\,\,\,\,\,x=-7-5 \,\,\, \text{ or } \,\,\, x=7-5\)
\(\,\,\,\,\,\,\)The answer is \( x=-12, x=2 \)

 

\(\textbf{2)}\) \( 2x^2+7x=4 \)

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The answer is \( x= -4, \frac{1}{2} \)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,2x^2+7x=4\)
\(\,\,\,\,\,\,x^2+\frac{7}{2}x=2\)
\(\,\,\,\,\,\,x^2+\frac{7}{2}x+\text{___}=2+\text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{\frac{7}{2}}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{7}{4}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\frac{49}{16}\)
\(\,\,\,\,\,\,x^2+\frac{7}{2}x+\underline{\frac{49}{16}}=2+\underline{\frac{49}{16}}\)
\(\,\,\,\,\,\,x^2+\frac{7}{2}x+\frac{49}{16}=\frac{81}{16}\)

\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x+\frac{7}{4})^2=\frac{81}{16}\)

\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{\left(x+\frac{7}{4}\right)^2}=\sqrt{\frac{81}{16}}\)
\(\,\,\,\,\,\,x+\frac{7}{4}=\pm\frac{9}{4}\)
\(\,\,\,\,\,\,x=\pm\frac{9}{4}-\frac{7}{4}\)
\(\,\,\,\,\,\,x=-\frac{9}{4}-\frac{7}{4}\,\,\, \text{ or } \,\,\, x=\frac{9}{4}-\frac{7}{4}\)
\(\,\,\,\,\,\,x=-\frac{16}{4}\,\,\, \text{ or } \,\,\, x=\frac{2}{4}\)
\(\,\,\,\,\,\,\)The answer is \( x=-4, , x=\frac{1}{2} \)

 

\(\textbf{3)}\) \( x^2-6x+8=0 \)

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The answer is \( x=2, x=4 \)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2-6x+8=0\)
\(\,\,\,\,\,\,x^2-6x=-8\)
\(\,\,\,\,\,\,x^2-6x+\text{___}=-8+\text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{-6}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(-3\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle9\)
\(\,\,\,\,\,\,x^2-6x+\underline{9}=-8+ \underline{9}\)
\(\,\,\,\,\,\,x^2-6x+9=1\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x-3)^2=1\)
\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x-3)^2}=\sqrt{1}\)
\(\,\,\,\,\,\,x-3=\pm1\)
\(\,\,\,\,\,\,x=\pm1+3\)
\(\,\,\,\,\,\,x=-1+3 \,\,\, \text{ or } \,\,\, x=1+3\)
\(\,\,\,\,\,\,\)The answer is \( x=2, x=4 \)

 

\(\textbf{4)}\) \(x^2+5x+6=0\)

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The answer is \(x=-3,-2\)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2+5x+6=0\)
\(\,\,\,\,\,\,x^2+5x=-6\)
\(\,\,\,\,\,\,x^2+5x+\text{___}=-6+ \text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{5}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\frac{25}{4}\)
\(\,\,\,\,\,\,x^2+5x+\underline{\frac{25}{4}}=-6+ \underline{\frac{25}{4}}\)
\(\,\,\,\,\,\,x^2+5x+6.25=.25\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x+2.5)^2=.25\)
\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x+2.5)^2}=\sqrt{.25}\)
\(\,\,\,\,\,\,x+2.5=\pm.5\)
\(\,\,\,\,\,\,x=\pm.5-2.5\)
\(\,\,\,\,\,\,x=-.5-2.5 \,\,\, \text{ or } \,\,\, x=.5-2.5\)
\(\,\,\,\,\,\,\)The answer is \( x=-3, x=-2 \)

 

\(\textbf{5)}\) \(x^2-6x+8=0\)

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The answer is \(x=2,4\)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2-6x+8=0\)
\(\,\,\,\,\,\,x^2-6x=-8\)
\(\,\,\,\,\,\,x^2-6x+{\text{___}}=-8+{\text{___}}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=\displaystyle\left(\frac{-6}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___}=9\)
\(\,\,\,\,\,\,x^2-6x+\underline{9}=-8+\underline{9}\)
\(\,\,\,\,\,\,x^2-6x+9=-8+9\)
\(\,\,\,\,\,\,(x-3)^2=1\)
\(\text{Step 3: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x-3)^2}=\pm\sqrt{1}\)
\(\,\,\,\,\,\,x-3=\pm1\)
\(\,\,\,\,\,\,x=3\pm1\)
\(\,\,\,\,\,\,x=3+1 \,\,\, \text{ or } \,\,\, x=3-1\)
\(\,\,\,\,\,\,\)The answer is \(x=4, x=2\)

 

\(\textbf{6)}\) \(x^2-8x=20\)

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The answer is \(x=-2,10\)

 

\(\textbf{7)}\) \( x^2 – 4x – 1 = 0 \)

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The answer is \( x = 2 + \sqrt{5}, x = 2 – \sqrt{5} \)

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\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2 – 4x – 1 = 0\)
\(\,\,\,\,\,\,x^2 – 4x = 1\)
\(\,\,\,\,\,\,x^2 – 4x + \text{___} = 1 + \text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{-4}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle 4\) \(\,\,\,\,\,\,x^2 – 4x + \underline{4} = 1 + \underline{4}\)
\(\,\,\,\,\,\,x^2 – 4x + 4 = 5\)
\(\text{Step 3: Complete the square.}\) \(\,\,\,\,\,\,(x – 2)^2 = 5\)
\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x – 2)^2} = \pm \sqrt{5}\)
\(\,\,\,\,\,\,x – 2 = \pm \sqrt{5}\)
\(\,\,\,\,\,\,x = 2 \pm \sqrt{5}\)
\(\,\,\,\,\,\,\)The answer is \( x = 2 + \sqrt{5}, x = 2 – \sqrt{5} \)

 

\(\textbf{8)}\) \( x^2 + 6x + 2 = 0 \)

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The answer is \( x = -3 + \sqrt{7}, x = -3 – \sqrt{7} \)

Show Work

\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2 + 6x + 2 = 0\)
\(\,\,\,\,\,\,x^2 + 6x = -2\)
\(\,\,\,\,\,\,x^2 + 6x + \text{___} = -2 + \text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{6}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle 9\)
\(\,\,\,\,\,\,x^2 + 6x + \underline{9} = -2 + \underline{9}\)
\(\,\,\,\,\,\,x^2 + 6x + 9 = 7\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x + 3)^2 = 7\)
\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x + 3)^2} = \pm \sqrt{7}\)
\(\,\,\,\,\,\,x + 3 = \pm \sqrt{7}\)
\(\,\,\,\,\,\,x = -3 \pm \sqrt{7}\)
\(\,\,\,\,\,\,\)The answer is \( x = -3 + \sqrt{7}, x = -3 – \sqrt{7} \)

 

\(\textbf{9)}\) \( x^2 – 2x – 3 = 0 \)

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The answer is \( x = 3, -1 \)

Show Work

\(\text{Step 1: Set up the blank spaces.}\)
\(\,\,\,\,\,\,x^2 – 2x – 3 = 0\)
\(\,\,\,\,\,\,x^2 – 2x = 3\)
\(\,\,\,\,\,\,x^2 – 2x + \text{___} = 3 + \text{___}\)
\(\text{Step 2: Plug } \left(\frac{b}{2}\right)^2 \text{ into the spaces.}\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{b}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle\left(\frac{-2}{2}\right)^2\)
\(\,\,\,\,\,\,\text{___} = \displaystyle 1\)
\(\,\,\,\,\,\,x^2 – 2x + \underline{1} = 3 + \underline{1}\)
\(\,\,\,\,\,\,x^2 – 2x + 1 = 4\)
\(\text{Step 3: Complete the square.}\)
\(\,\,\,\,\,\,(x – 1)^2 = 4\)
\(\text{Step 4: Solve for x.}\)
\(\,\,\,\,\,\,\sqrt{(x – 1)^2} = \pm \sqrt{4}\)
\(\,\,\,\,\,\,x – 1 = \pm 2\)
\(\,\,\,\,\,\,x = 1 \pm 2\)
\(\,\,\,\,\,\,x = 1 + 2 \,\,\, \text{or} \,\,\, x = 1 – 2\)
\(\,\,\,\,\,\,\)The answer is \( x = 3, x = -1 \)

 

 

See Related Pages\(\)

\(\bullet\text{ Adding and Subtracting Polynomials}\)
\(\,\,\,\,\,\,\,\,(4d+7)−(2d−5)…\)

\(\bullet\text{ Multiplying Polynomials}\)
\(\,\,\,\,\,\,\,\,(x+2)(x^2+3x−5)…\)

\(\bullet\text{ Dividing Polynomials}\)
\(\,\,\,\,\,\,\,\,(x^3-8)÷(x-2)…\)

\(\bullet\text{ Dividing Polynomials (Synthetic Division)}\)
\(\,\,\,\,\,\,\,\,(x^3-8)÷(x-2)…\)

\(\bullet\text{ Synthetic Substitution}\)
\(\,\,\,\,\,\,\,\,f(x)=4x^4−3x^2+8x−2…\)

\(\bullet\text{ End Behavior}\)
\(\,\,\,\,\,\,\,\, \text{As } x\rightarrow \infty, \quad f(x)\rightarrow \infty \)
\(\,\,\,\,\,\,\,\, \text{As } x\rightarrow -\infty, \quad f(x)\rightarrow \infty… \)

\(\bullet\text{ Completing the Square}\)
\(\,\,\,\,\,\,\,\,x^2+10x−24=0…\)

\(\bullet\text{ Quadratic Formula and the Discriminant}\)
\(\,\,\,\,\,\,\,\,x=-b \pm \displaystyle\frac{\sqrt{b^2-4ac}}{2a}…\)

\(\bullet\text{ Complex Numbers}\)
\(\,\,\,\,\,\,\,\,i=\sqrt{-1}…\)

\(\bullet\text{ Multiplicity of Roots}\)
\(\,\,\,\,\,\,\,\,\)\(…\)

\(\bullet\text{ Rational Zero Theorem}\)
\(\,\,\,\,\,\,\,\, \pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12…\)

\(\bullet\text{ Descartes Rule of Signs}\)
\(\,\)

\(\bullet\text{ Roots and Zeroes}\)
\(\,\,\,\,\,\,\,\,\text{Solve for }x. 3x^2+4x=0…\)

\(\bullet\text{ Linear Factored Form}\)
\(\,\,\,\,\,\,\,\,f(x)=(x+4)(x+1)(x−3)…\)

\(\bullet\text{ Polynomial Inequalities}\)
\(\,\,\,\,\,\,\,\,x^3-4x^2-4x+16 \gt 0…\)

 

In Summary

Completing the square is a mathematical technique used to solve quadratic equations. It involves taking a quadratic equation and re-arranging it into the form of a perfect square trinomial, which makes it easier to solve. Completing the square is a useful method because it allows us to solve quadratic equations that may not have an obvious solution using other methods. It is also a useful tool for graphing quadratic equations because it gives us the coordinates of the vertex of the parabola.

Completing the square is typically introduced in algebra 2 or precalculus classes.

The most common mistakes when completing the square are related to the steps involving the \(\left(\frac{b}{2}\right)^2\). But with practice, these mistakes show up less.

René Descartes and Isaac Newton both used completing the square to solve quadratic equations and understand the properties of quadratic functions. Today, completing the square is still a useful mathematical tool.

Topics related to completing the square include:

Quadratic equations: Completing the square is a technique used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). Solving quadratic equations is an important skill in algebra, and there are several different methods for doing so, including factoring, using the quadratic formula, and graphing.

Parabolas: Completing the square is a useful tool for understanding the properties of parabolas, which are the graphs of quadratic functions. By completing the square, we can find the vertex form of the parabola, which quickly shows the coordinates of the vertex, the shape and the direction of the graph.

Factoring: Completing the square is closely related to factoring quadratic expressions. In fact, the technique of completing the square can be thought of as a way to factor a quadratic expression into the form (x + p)^2 = q, where p and q are constants. Factoring is a fundamental skill in algebra, and it is used in many different contexts, including solving equations and simplifying expressions.

The quadratic formula: The quadratic formula is another method for solving quadratic equations. It states that the solutions to the equation \(ax^2 + bx + c = 0\) are given by the formula \(x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\). The quadratic formula can be derived using completing the square.