Chapter 1.8, Problem 67E

Solution

$\left(a\right)$

To find: The solution of the equation $\left|x^2-3x-14\right|=4$ .

The solution of the equation is $x=\left\{-3,-2,5,6\right\}$ .

Given:

The equation is $\left|x^2-3x-14\right|=4$ .

Concept used:

For the absolute value function, if $\left|f\left(x\right)\right|=a$ we have,

  $f\left(x\right)=\left\{\begin{array}{l}a\text{, for }a\ge 0\\ -a,\text{for }a<0\end{array}\right.$

Calculation:

With the help of absolute value function, the equation $\left|x^2-3x-14\right|=4$ is re-written as:

  $\begin{array}{c}{x}^2-3x-14=4\text{ or }{x}^2-3x-14=-4\\ x^2-3x-18=0\text{ or }{x}^2-3x-10=0\\ x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)\left(-18\right)}}{2\left(1\right)}\text{ or }x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(1\right)\left(-10\right)}}{2\left(1\right)}\\ x=\frac{3\pm \sqrt{81}}{2}\text{ or }x=\frac{3\pm \sqrt{49}}{2}\\ x=\frac{3\pm 9}{2}\text{ or }x=\frac{3\pm 7}{2}\\ x=\frac{3+9}{2}\text{,}\frac{3-9}{2}\text{ or }x=\frac{3+7}{2},\frac{3-7}{2}\\ x=6\text{,}-3\text{ or }x=5,-2\\ x=\left\{-3,-2,5,6\right\}\end{array}$

Hence, the required solution of equation is $x=\left\{-3,-2,5,6\right\}$ .

$\left(b\right)$

To find: The solution of the equation $x^2=\left|x\right|+6$ .

The solution of the equation is $x=\left\{-3,3\right\}$ .

Given:

The equation is $x^2=\left|x\right|+6$ .

Concept used:

For the absolute value function, if $\left|f\left(x\right)\right|=a$ we have,

  $f\left(x\right)=\left\{\begin{array}{l}a\text{, for }a\ge 0\\ -a,\text{for }a<0\end{array}\right.$

Calculation:

With the help of absolute value function, the equation $x^2=\left|x\right|+6$ is re-written as:

  $\begin{array}{c}{x}^2=x+6\text{ or }{x}^2=-x+6\\ x^2-x-6=0\text{ or }{x}^2+x-6=0\\ x=\frac{-\left(-1\right)\pm \sqrt{{\left(-1\right)}^2-4\left(1\right)\left(-6\right)}}{2\left(1\right)}\text{ or }x=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(1\right)\left(-6\right)}}{2\left(1\right)}\\ x=\frac{1\pm \sqrt{25}}{2}\text{ or }x=\frac{-1\pm \sqrt{25}}{2\left(1\right)}\\ x=\frac{1\pm 5}{2}\text{ or }x=\frac{-1\pm 5}{2}\\ x=\frac{1+5}{2}\text{,}\frac{1-5}{2}\text{ or }x=\frac{-1+5}{2},\frac{-1-5}{2}\\ x=3\text{,}-2\text{ or }x=2,-3\\ x=\left\{-3,-2,2,3\right\}\end{array}$

Now, the solutions $x=-2\text{and }x=2$ doesn’t satisfy the equation $x^2=\left|x\right|+6$ . Therefore, $x=-2\text{and }x=2$ will not be included in solution set and the solutions $x=-3\text{and }x=3$ satisfy the equation $x^2=\left|x\right|+6$ . So, $x=-3\text{and }x=3$ will be included in solution set.

Hence, the required solution of equation is $x=\left\{-3,3\right\}$ .