Chapter 9, Problem 34CR

To determine

To solve the system using any method.

Answer to Problem 34CR

The solutions are $x=1-\sqrt{15},x=1+\sqrt{15}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}y=x^2-9\\ y=2x+5\end{array}$

Calculation:

Let the equation be (1) and (2).

Subtracting (2) from (1),

  $x^2-9-2x-5=y-y$

Solving,

  $x^2-2x-14=0$

Here,

  $a=1,b=-2,c=-14$

Using quadratic formula,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting values,

  $x=\frac{-(-2)\pm \sqrt{{(-2)}^2-4(1)(-14)}}{2(1)}$

Solving,

  $x=\frac{2\pm \sqrt{4+56}}{2}$

So,

  $x=\frac{2\pm \sqrt{60}}{2}$

The solutions are:

  $x=\frac{2-\sqrt{60}}{2}x=\frac{2+\sqrt{60}}{2}$

Solving,

  $x=\frac{2-2\sqrt{15}}{2}x=\frac{2+2\sqrt{15}}{2}$

Taking out common,

  $x=\frac{2(1-\sqrt{15})}{2}x=\frac{2(1+\sqrt{15})}{2}$

Hence, the solutions are $x=1-\sqrt{15},x=1+\sqrt{15}$ .