Chapter 9.5, Problem 20E

To determine

To solve the equation using Quadratic formula.

Answer to Problem 20E

  $x=\frac{2+\sqrt{14}}{5}$ and $x=\frac{2-\sqrt{14}}{5}$

Explanation of Solution

Given:

Equation:

  $5x^2-2=4x$

Calculation:

Converting the equation in standard form:

  $5x^2-4x-2=0$

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

  $a=5,b=-4,c=-2$

Finding the value of $x$

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substituting the values,

  $x=\frac{-(-4)\pm \sqrt{{(-4)}^2-4(5)(-2)}}{2(5)}$

Opening brackets,

  $x=\frac{4\pm \sqrt{16+40}}{10}$

So,

  $x=\frac{4\pm \sqrt{56}}{10}$

Solving root,

  $x=\frac{4\pm 2\sqrt{14}}{10}$

Taking common,

  $x=\frac{2(2\pm \sqrt{14})}{10}$

So,

  $x=\frac{2\pm \sqrt{14}}{5}$

Therefore,

  $x=\frac{2+\sqrt{14}}{5}$ and $x=\frac{2-\sqrt{14}}{5}$