Chapter A.5, Problem 92E

To determine

Verify the equation.

Answer to Problem 92E

  $\boxed{\left[-1,15\right]}$

Explanation of Solution

Given information:

Solve the equation. Check your solutions.

  $\left|x-15\right|=x^2-15x$

Calculation:

Here, we will consider the following equation:

  $\left|x-15\right|=x^2-15x$

Now, by absolute principal value for any positive number $p$ and algebraic expression $X$ , the solution of $\left|X\right|=p$ are those numbers that satisfy $X=-p$ or $X=p$ .

Thus,

  $\begin{array}{l}X=x-15\\ p=x^2-15x\end{array}$

Now, the two equations are as follows:

  $\begin{array}{l}x-15=x^2-15x\\ x-15=-\left(x^2-15x\right)\end{array}$

We will first solve the equation:

  $x-15=x^2-15x$

Now, subtract $x$ from both sides:

  $x-15-x=x^2-15x-x$

Now, combine the like terms:

  $-15=x^2-16x$

Now, add $15$ to both sides:

  $-15+15=x^2-16+15$

Now, again combine the like terms:

  $0=x^2-16x+15$

If $a=b$ then $b=a$

Thus, the above equation can be written as:

  $x^2-16x+15=0$

Now, factor the above expression:

  $\left(x-1\right)\left(x-15\right)=0$

Now, we will use the zero factor property for the values of $x$ are as follows:

  $x=1,15$

Here, we will consider the equation:

  $x-15=-\left(x^2-15x\right)$

Now, distribute $-1$ into the bracket:

  $x-15=-x^2+15x$

Now, subtract $x$ from both the sides:

  $x-15-x=-x^2+15x-x$

Now, combine the like terms:

  $-15=-x^2+14x$

Now, add $15$ to both sides:

  $-15+15=-x^2+14x+15$

Now, again combine the like terms:

  $0=-x^2-14x+15$

Now, multiply both sides by $-1$ :

  $0=x^2-14x-15$

If $a=b$ then $b=a$

Thus, above expression can be written as:

  $x^2-14x-15=0$

Now, factor the equation:

  $\left(x+1\right)\left(x-15\right)=0$

By using zero factor property:

  $x=-1,15$

Now, to verify the equation we will put $x=-1$ in the equation $\left|x-15\right|=x^2-15x$ .

  $\begin{array}{l}\left|-1-15\right|={\left(-1\right)}^2-15\times \left(-1\right)\\ \left|-16\right|=16\\ 16=16\end{array}$

It is true.

Now, put $x=15$ in the equation $\left|x-15\right|=x^2-15x$ :

  $\begin{array}{l}\left|15-15\right|={\left(15\right)}^2-15\times 15\\ 0=0\end{array}$

It is true.

Now, put $x=1$ in the equation $\left|x-15\right|=x^2-15x$ :

  $\begin{array}{l}\left|1-15\right|\ne {\left(1\right)}^2-15\left(1\right)\\ 14\ne -14\end{array}$

It is not true. Thus, we will discard the value $x=1$ .

Hence, the solution set is $\boxed{\left[-1,15\right]}$ .