Chapter 9.6, Problem 38E

To determine

To calculate the solution of the given set of equations.

Answer to Problem 38E

The solution is $x\approx 0.31,2.31$ and $y\approx -3.3,0.31$

Explanation of Solution

Given information:

  $\begin{array}{l}y=2x^2+x-8\\ y=x^2-5\end{array}$

Formula used:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

Equation 1 is $y=2x^2+x-8$

Equation 2 is $y=x^2-5$

Graphing the two equations,

  

Concluding from the graph, the set of equation has two solution, which lines between $x=0$ and $x=1$ ,

  $x=-2$ and $x=-3$ ,

Substituting equation 2 in equation 1,

  $2x^2+x-8=x^2-5$

Shifting the values,

  $2x^2+x-8-x^2+5=0$

On solving,

  $x^2+x-3=0$

Hence,

  $a=1$ ,

  $b=1$ and

  $c=-3$

Hence, substituting the values,

  $x=\frac{-1\pm \sqrt{1^2-4(1)(-3)}}{2(1)}$

So,

  $x=\frac{-1\pm \sqrt{1+12}}{2}$

On solving,

  $x=\frac{-1\pm 3.61}{2}$

For +3.61,

  $x=\frac{-1+3.61}{2}$

On solving,

  $x=1.31$

Hence for y, substituting the values,

  $y={(1.31)}^2-5$

On solving,

  $y=-3.3$

For -3.61,

  $x=\frac{-1-3.61}{2}$

On solving,

  $x=2.31$

Hence for y, substituting the values,

  $y={(2.31)}^2-5$

On solving,

  $y=0.31$

Hence, the solution is $x\approx 0.31,2.31$ and $y\approx -3.3,0.31$