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$x$.

$6x^2=5x+6$

Short answer

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

Step-by-step solution

Step 1. 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.

Step 2. Substitute the values.

It is given that $6x^2=5x+6$which can also be written as $6x^2-5x-6=0$. Using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 6 for a, $-5$ for b and $-6$ for c.

Put the values in the formula and get $x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\times \left(6\right)\times \left(-6\right)}}{2\left(6\right)}$.

Step 3. Simplify for x.

Simplify the expression for x.

$$\begin{gathered} x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\times \left(6\right)\times \left(-6\right)}}{2\left(6\right)} \\ =\frac{5\pm \sqrt{25+144}}{12} \\ =\frac{5\pm \sqrt{169}}{12} \\ =\frac{5\pm 13}{12} \end{gathered}$$

Therefore, either $x=\frac{5+13}{12}=\frac{18}{12}$or $x=\frac{5-13}{12}=\frac{-8}{12}$that is., either $x=\frac{3}{2}$or $x=-\frac{2}{3}$. Here, the $x$is having two values $\frac{3}{2}$and $-\frac{2}{3}$. Since, $x$is a length of a segment that could not be negative or zero, therefore, the value of $x$is $\frac{3}{2}$.