Chapter 4, Problem 31AR

To determine

To solve the given equation, $2x^2+x-3=0$ by factoring or quadratic formula where x represents the length of a segment.

Answer to Problem 31AR

  $x=1$

Explanation of Solution

Given:

The equation is, $2x^2+x-3=0$

Calculation:

We can solve the given equation by using quadratic formula.

That is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Now, the equation $2x^2+x-3=0$ , which is in the form of $a{x}^2+bx+c=0$ . Where, $a=2,b=1\text{ and c =-3}$

Now, Substitute the values in the formula then we get that,

  $\begin{array}{l}x=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\left(2\right)\left(-3\right)}}{2\left(2\right)}\\ x=\frac{-1\pm \sqrt{1+24}}{4}\\ x=\frac{-1\pm \sqrt{25}}{4}\\ x=\frac{-1\pm 5}{4}\\ x=\frac{-1+5}{4},\frac{-1-5}{4}\\ x=\frac{4}{4},\frac{-6}{4}\\ x=1,\frac{-3}{2}\end{array}$

Here, the x is having two values. $x=1$ and $x=\frac{-3}{2}$ . In the question given that x is a length of a segment so x should not be negative.

Therefore, the value of x is 1.