Question

In Exercises 28-33 $x$represents the length of a segment. When a value of $x$doesn't make sense as a length, eliminate that value of$x$.

$2x^2+x-3=0$

Short answer

The length of the segment is$x=1$.

Step-by-step solution

- Understand the concept used 

If $a{x}^2+bx+c=0$, with $a\ne 0$, then $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$. Solving using factoring can be done by splitting the middle term of equation and then making groups and equating them to zero.

- Substitute the values 

It is given that$2x^2+x-3=0$ .Using quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 2 for a, 1 for b and -3 for c.

Put the values in the formula and get$x=\frac{-\left(1\right)\pm \sqrt{{\left(1\right)}^2-4\times \left(2\right)\times \left(-3\right)}}{2\left(2\right)}$ .

- Simplify for x 

Simplify the expression for x.

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

Therefore, either $x=\frac{-1+5}{4}=\frac{4}{4}$or $x=\frac{-1-5}{4}=\frac{-6}{4}$that is., either $x=1$or $x=-\frac{3}{2}$. Here, the $x$is having two values 1 and $-\frac{3}{2}$. Since$x$, is a length of a segment that could not be negative or zero, therefore, the value of $x$is 1.