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.

$x\left(x-50\right)=0$

Short answer

The length of the segment is$x=50$

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$x(x-50)=0$which can also be written as$x^2-50x=0$ .Using quadratic formula$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, substitute 1 for a, -50 for b and 0 for c.

Put the values in the formula and get $x=\frac{-\left(-50\right)\pm \sqrt{{\left(-50\right)}^2-4\times \left(1\right)\times \left(0\right)}}{2\left(1\right)}$.

- Simplify for x 

Simplify the expression for x.

$\begin{array}{c}x=\frac{-\left(-50\right)\pm \sqrt{{\left(-50\right)}^2-4\times \left(1\right)\times \left(0\right)}}{2\left(1\right)}\\ =\frac{50\pm \sqrt{2500}}{2}\\ =\frac{50\pm 50}{2}\end{array}$

Therefore, either $x=\frac{50+50}{2}=\frac{100}{2}$or $x=\frac{50-50}{2}$that is., either $x=50$or $x=0$. Here, the is having two values 50 and 0. Since , x is a length of a segment that could not be negative or zero, therefore, the value of x is 50.