Chapter 4, Problem 28AR

To determine

To solve the given equation, $x\left(x-5\right)=0$ by factoring or quadratic formula where x represents the length of a segment.

Answer to Problem 28AR

  $x=5$

Explanation of Solution

Given:

The equation is, $x\left(x-5\right)=0$

Calculation:

We can solve the given equation by using quadratic formula.

That is,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

We can write the given equation $x\left(x-5\right)=0$ as

  $\begin{array}{l}x\left(x-5\right)=0\\ x^2-5x=0\end{array}$

Now, the equation $x^2-5x=0$ , which is in the form of $a{x}^2+bx+c=0$ . Where, $a=1,b=-5\text{ and c =0}$

Now, Substitute the values in the formula then we get that,

  $\begin{array}{l}x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}\\ x=\frac{5\pm \sqrt{25-0}}{2}\\ x=\frac{5\pm 5}{2}\\ x=\frac{5+5}{2},\frac{5-5}{2}\\ x=5\text{ and }0\end{array}$

Here, the x is having two values. 0 and 5. In the question given that x is a length of a segment that could not be zero.

Therefore, the value of x is 5.