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 $x$.

$2x^2-7x-4=0$

Short answer

The length of the segment is$x=4$ .

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$2x^2-7x-4=0$.Using quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 2 for a, -7 for b and -4 for c.

Put the values in the formula and get$x=\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\times \left(2\right)\times \left(-4\right)}}{2\left(2\right)}$ .

- Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-\left(-7\right)\pm \sqrt{{\left(-7\right)}^2-4\times \left(2\right)\times \left(-4\right)}}{2\left(2\right)}\\ =\frac{7\pm \sqrt{49+32}}{4}\\ =\frac{7\pm \sqrt{81}}{4}\\ =\frac{7\pm 9}{4}\end{array}$

Therefore, either $x=\frac{7+9}{4}=\frac{16}{4}$or $x=\frac{7-9}{4}=\frac{-2}{4}$that is., either $x=4$or$x=-\frac{1}{2}$ . Here, the $x$is having two values 4 and $-\frac{1}{2}$. Since, $x$is a length of a segment that could not be negative or zero, therefore, the value of $x$is 4.