Chapter 9.6, Problem 30PPS

To determine

Solve the equation.

Answer to Problem 30PPS

Quadratic Formula

Explanation of Solution

Given Information: $3x^2-24x=-36$

Formula Used: $D=b^2-4ac$ and $x=\frac{-b\pm \sqrt{D}}{2a}$

Calculation:

• $D$ equal to negative so $x$ has no real solution.
• $D$ equal to 0 so $x$ has 1 real solution.
• $D$ equal to positive so $x$ has 2 real solution.

  $3x^2-24x=-36$

So, simplify

  $3x^2-24x+36=0$

Now, use formula $D=b^2-4ac$ to check if solution of $x$ exist or not.

As $a=3,b=-24$ and $c=36$

  $D={\left(-24\right)}^2-4\times 3\times 36$

Now,

  $D=576-432$

Hence,

  $D=144$

As, $D$ equal to positive so $x$ has 2 real solution.

  $x=\frac{-b\pm \sqrt{D}}{2a}$

So,

  $x=\frac{-\left(-24\right)\pm \sqrt{144}}{2\times 3}$

Now, square root of 144 is 12.

  $x=\frac{24\pm 12}{6}$

Now,

$x=\frac{24+12}{6}$ or $x=\frac{24-12}{6}$

Now,

$x=\frac{36}{6}$ or $x=\frac{12}{6}$

Hence,

$x=6$ or $x=2$