Chapter 9.5, Problem 15E

To determine

To solve the equation using Quadratic formula.

Answer to Problem 15E

  $x=\frac{3}{2}$ and $x=\frac{2}{3}$

Explanation of Solution

Given:

Equation:

  $6x^2-13x=-6$

Calculation:

Converting the equation in standard form:

  $6x^2-13x+6=0$

On comparing the equation with $a{x}^2+bx+c$

  $a=6,b=-13,c=6$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values,

  $x=\frac{-(-13)\pm \sqrt{{(-13)}^2-4(6)(6)}}{2(6)}$

Opening brackets,

  $x=\frac{13\pm \sqrt{169-144}}{12}$

So,

  $x=\frac{13\pm \sqrt{25}}{12}$

Solving root,

  $x=\frac{13\pm 5}{12}$

So,

  $x=\frac{13+5}{12}$ and $x=\frac{13-5}{12}$

Therefore,

  $x=\frac{18}{12}$ and $x=\frac{8}{12}$

So,

  $x=\frac{3}{2}$ and $x=\frac{2}{3}$