Question

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

$x^2-17x+72=0$

Short answer

The length of the segment is$x=9orx=8$.

Step-by-step solution

- Understand the concept used 

$ifa{x}^2+bx+c=0$$witha\ne 0$$thenx=\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 $x^2-17x+72=0$Using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -17 for b and 72 for c.

Put the values in the formula and get$x=\frac{-\left(-17\right)\pm \sqrt{{\left(-17\right)}^2-4\times \left(1\right)\times \left(72\right)}}{2\left(1\right)}$

- Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-\left(-17\right)\pm \sqrt{{\left(-17\right)}^2-4\times \left(1\right)\times \left(72\right)}}{2\left(1\right)}\\ =\frac{17\pm \sqrt{289-288}}{2}\\ =\frac{17\pm \sqrt{1}}{2}\\ =\frac{17\pm 1}{2}\end{array}$

Therefore, either$x=\frac{17+1}{2}=\frac{18}{2}$or$x=\frac{17-1}{2}=\frac{16}{2}$that is., either$x=9$or$x=8$ Here, the x is having two values 9 and 8. Since, x is a length of a segment that could not be negative or zero, therefore, the value x of can be either 9 or 8.