Chapter 9, Problem 5QC

To determine

To find: The solution by using completing the square.

Answer to Problem 5QC

The solution is $x=\frac{3}{2}$ and $x=-5$ .

Explanation of Solution

Given information:

The given equation is $2x^2+7x-15=0$ .

Calculation:

Isolate the variable on one side and the constant terms on the other.

  $\begin{array}{c}2x^2+7x-15=0\\ 2x^2+7x=15\end{array}$

Make coefficient of $x^2$ term $1$ by dividing both side of the equation by $2$ .

  $\begin{array}{c}\frac{2x^2+7x}{2}=\frac{15}{2}\\ \frac{2x^2}{2}+\frac{7x}{2}=\frac{15}{2}\\ x^2+\frac{7}{2}x=\frac{15}{2}\end{array}$

Calculate value of $b$ , the coefficient of $x$ .

Comparing the left side of equation with $x^2+bx$ .

  $b=\frac{7}{2}$

The coefficient of $x$ is $\frac{7}{2}$ .

Calculate the value of ${\left(\frac{1}{2}b\right)}^2$ .

  $\begin{array}{c}{\left(\frac{1}{2}b\right)}^2={\left(\frac{1}{2}\cdot \frac{7}{2}\right)}^2\\ ={\left(\frac{7}{4}\right)}^2\\ =\frac{49}{16}\end{array}$

The number to calculate the square is $\frac{49}{16}$ .

Add $\frac{49}{16}$ on both side of the equation.

  $\begin{array}{l}{x}^2+\frac{7}{2}x+\frac{49}{16}=\frac{15}{2}+\frac{49}{16}\\ x^2+\frac{7}{2}x+\frac{49}{16}=\frac{169}{16}\end{array}$

Now factor the perfect square trinomial, writing it as a binomial square.

  $\begin{array}{c}{x}^2+2\cdot x\cdot \frac{7}{4}+{\left(\frac{7}{4}\right)}^2=\frac{169}{16}\\ {\left(x+\frac{7}{4}\right)}^2=\frac{169}{16}\end{array}$

The perfect square trinomial is ${\left(x+\frac{7}{4}\right)}^2=\frac{169}{16}$ .

Using the square root properly.

  $\begin{array}{l}x+\frac{7}{4}=\pm \sqrt{\frac{169}{16}}\\ x+\frac{7}{4}=\pm \frac{13}{4}\end{array}$

Calculate the value of $x$ .

  $\begin{array}{c}x+\frac{7}{4}=\frac{13}{4}\\ x=\frac{13}{4}-\frac{7}{4}\\ x=\frac{6}{4}\\ x=\frac{3}{2}\end{array}$

  $\begin{array}{c}x+\frac{7}{4}=-\frac{13}{4}\\ x=-\frac{13}{4}-\frac{7}{4}\\ x=-\frac{20}{4}\\ x=-5\end{array}$

The solution is $x=\frac{3}{2}$ and $x=-5$ .

Verify the solution by replacing the value of $x$ by $\frac{3}{2}$ and $-5$ .

  $\begin{array}{c}2{\left(\frac{3}{2}\right)}^2+7\left(\frac{3}{2}\right)-15=0\\ \frac{9}{2}+\frac{21}{2}-15=0\\ 15-15=0\\ 0=0\end{array}$

  $\begin{array}{c}2{\left(-5\right)}^2+7\left(-5\right)-15=0\\ 50-35-15=0\\ 15-15=0\\ 0=0\end{array}$

The solution is verified.

Therefore, the solution is $x=\frac{3}{2}$ and $x=-5$ .