Chapter 9.6, Problem 26PPS

To determine

The solutions of the equation $-3x^2=8x-12$ using quadratic formula

Answer to Problem 26PPS

  $x=\frac{-4}{3}+\frac{2\sqrt{13}}{3},\text{and }x=\frac{-4}{3}-\frac{2\sqrt{13}}{3}$

Explanation of Solution

Given:

The equation,

  $-3x^2=8x-12$

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

In order to solve the given equation $-3x^2-8x+12=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=-3,b=-8,\text{ and }c=12$

Now, above quadratic equation $-3y^2-8y+12=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-8\right)\pm \sqrt{{\left(-8\right)}^2-4\left(-3\right)\left(12\right)}}{2\left(-3\right)}\\ =\frac{8\pm \sqrt{64+144}}{-6}\\ =\frac{8\pm \sqrt{208}}{-6}\\ =\frac{8\pm \sqrt{16\times 13}}{-6}\\ =\frac{8\pm 4\sqrt{13}}{-6}\\ =\frac{2\left(4\pm 2\sqrt{13}\right)}{-6}\\ =\frac{4\pm 2\sqrt{13}}{-3}\\ x=\frac{4+2\sqrt{13}}{-3},\text{and }x=\frac{4-2\sqrt{13}}{-3}\\ x=\frac{-4}{3}+\frac{2\sqrt{13}}{3},\text{and }x=\frac{-4}{3}-\frac{2\sqrt{13}}{3}\end{array}$

Thus, the solutions of the given question are, $x=\frac{-4}{3}+\frac{2\sqrt{13}}{3},\text{and }x=\frac{-4}{3}-\frac{2\sqrt{13}}{3}$ .