Chapter 4: Q. 30 (page 163)

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.
$x^{2} - 17 x + 72 = 0$

Short Answer

Expert verified

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

Step by step solution

- Understand the concept used

$i f a x^{2} + b x + c = 0$ $w i t h a \neq 0$ $t h e n x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ .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 $x^{2} - 17 x + 72 = 0$ Using quadratic formula $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ , substitute 1 for a, -17 for b and 72 for c.

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

- Simplify for x

Simplify the expression for x.

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

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.