Chapter ISG, Problem 8.5.4P

To determine

To solve the system of equations

Answer to Problem 8.5.4P

  $x=\frac{-1}{2}$ , -8 are the zeroes of equation

Explanation of Solution

Given information:

  $3x^2+24x-1.5x-12=0$

Calculation:

The given equation is $3x^2+24x-1.5x-12=0$

Solving,

  $3x^2+22.5x-12=0$

Dividing by 3,

  $x^2+7.5x-4=0$

Multiplying equation by 2,

  $2x^2+15x-8=0$

Now,

  $x^2+16x-x-8=0$

So,

  $2x(x+8)-(x+8)=0$

Taking common,

  $(2x-1)(x+8)=0$

For equation to be 0,

  $(2x-1)=0$ or $(x+8)=0$

Solving,

  $2x=-1,x=-8$

And,

  $x=\frac{-1}{2},x=-8$

Hence, $x=\frac{-1}{2}$ , -8 are the zeroes of equation