Chapter 9.5, Problem 20PPS

To determine

The solutions of the equation $2x^2-5x=-7$ using quadratic formula

Answer to Problem 20PPS

  $x=2+\frac{\sqrt{31}}{4},\text{ and }x=2-\frac{\sqrt{31}}{4}$

Explanation of Solution

Given:

The equation,

  $2x^2-5x=-7$

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 $2x^2-5x+7=0$ , 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

  $a=2,b=-5,\text{ and }c=7$

Now, above quadratic equation $2y^2-5y+7=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\left(2\right)\left(7\right)}}{2\left(2\right)}\\ =\frac{5\pm \sqrt{25-56}}{4}\\ =\frac{8\pm \sqrt{-31}}{4}\\ =\frac{8\pm \sqrt{31}}{4}\\ =\frac{8\pm \sqrt{31}}{4}\\ x=\frac{8}{4}+\frac{\sqrt{31}}{4},\text{ and x=}\frac{8}{4}-\frac{\sqrt{31}}{4}\\ \Rightarrow x=2+\frac{\sqrt{31}}{4},\text{ and }x=2-\frac{\sqrt{31}}{4}\end{array}$

Thus, the solutions of the given question are, $x=2+\frac{\sqrt{31}}{4},\text{ and }x=2-\frac{\sqrt{31}}{4}$ .