Chapter 12.6, Problem 4ST

To determine

To find: the value of an algebraic equation as mentioned in the problem.

Answer to Problem 4ST

The value of $3x^2+9x=0$ is $x=0,-3$ .

Explanation of Solution

Given information:

The equation mentioned is $3x^2+9x=0$ .

Formula:

The quadratic equation formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a,b,c$ is any constant.

Calculation:

Consider an algebraic equation as $3x^2+9x=0$ .

The mentioned equation is in the form of $a{x}^2+bx+c=0$ where $a=3,b=9,c=0$ .

Recall that quadratic equation formula $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a,b,c$ is any constant.

Therefore

  $\begin{array}{l}3x^2+9x=0\Rightarrow x=\frac{-9\pm \sqrt{9^2-4*1*0}}{2*3}\Rightarrow x=\frac{-9\pm \sqrt{81}}{6}\Rightarrow x=\frac{-9\pm 9}{6}\\ \Rightarrow x=\frac{-9+9}{6},\frac{-9-9}{6}\Rightarrow x=0,-3.\end{array}$

Therefore the value of $3x^2+9x=0$ is $x=0,-3$ .