Chapter 12, Problem 3CT

To determine

To calculate: The value of $\text{c}$ from the equation ${\text{c}}^{\text{2}}\text{+2c}-15=0$

Answer to Problem 3CT

The values are $\text{c}\text{=}\text{3}\text{or}\text{ c}\text{=}-\text{5}$ .

Explanation of Solution

Given information:

The expression is given as:

  ${\text{c}}^{\text{2}}\text{+2c}-15=0$

Formula used:

Method of completing the square;

For $x^2+\text{b}x+\underset{\_}{?}$

Step 1: Find half the coefficient of $x$ : $\frac{\text{b}}{\text{2}}$

Step 2: Square the result of step 1: ${\left(\frac{\text{b}}{\text{2}}\right)}^2$

Step 3: Add the result of step 2 to $x^2+\text{b}x$ : $x^2+\text{b}x+{\left(\frac{\text{b}}{\text{2}}\right)}^2$

Step 4: Completed the square: $x^2+\text{b}x+{\left(\frac{\text{b}}{\text{2}}\right)}^2={\left(x+\frac{\text{b}}{\text{2}}\right)}^2$

Calculation:

Apply the method of completing square,

By step 1, half of the coefficient of $x$ in equation ${\text{c}}^{\text{2}}\text{+2c}-15=0$ is $\frac{2}{2}=1$

By step 2, square the result of step 1 is 1.

Add 15 on the both sides of the equation as,

  $\begin{array}{c}{\text{c}}^{\text{2}}\text{+2c}-15+15=15\\ {\text{c}}^{\text{2}}\text{+2c}=15\end{array}$

By step 3 add the result of step 2 to both sides of equation,

  $\begin{array}{l}{\text{c}}^{\text{2}}\text{+2c+1}=15+1\\ {\text{c}}^{\text{2}}\text{+2c+1}=16\end{array}$

By step 4 compete the square ${\text{c}}^{\text{2}}\text{+2c+1}={\left(c+1\right)}^2$

Put in the equation,

  ${\left(\text{c+1}\right)}^2=16$

Take under root on both sides of equation,

  $\sqrt{{\left(\text{c+1}\right)}^2}=\pm \sqrt{16}$

Furthermore,

  $\begin{array}{c}\left(\text{c+1}\right)=\pm 4\\ c=\pm 4-1\end{array}$

The value of $\text{c}\text{=}\text{3}\text{or}\text{ c}\text{=}-\text{5}$ is $$ .