Chapter 9, Problem 38SGR

To determine

To solve the given equation by completing the square.

Answer to Problem 38SGR

The values of a are 0 and -10 .

Explanation of Solution

Given information :

The function provided is $-a^2-10a+25=25$ .

Formula used :

To solve an equation by completing the square, divide the coefficient of a of the equation by 2 and square it. That should provide the constant value. Then the equation would be of the form ${\left(x+\frac{b}{2}\right)}^2$ . Adjust the other constants and then square root the left-hand and right-hand sides. This would give the values of the equation.

Calculation :

  $-a^2-10a+25=25$

Given equation

  $\frac{-a^2-10a}{-1}=\frac{0}{-1}$

Subtracting 25 from both sides and dividing the equation by -1.

  $=>a^2+10a=0$

Equation for which the squares need to be found.

Here $b=10$ , the coefficient of a.

  $\frac{b}{2}=\frac{10}{2}$

  $=5$

For making the expression a perfect square.

  ${\left(\frac{b}{2}\right)}^2={\left(5\right)}^2$

  $=25$

This is the value of the constant.

Make left hand side a perfect square using the value of b.

  $=>a^2+10a+25=25$

Adjust the values accordingly

  $=>{\left(a+5\right)}^2=25$

Taking square root.

  $=>{\left(a+5\right)}^2=\sqrt{25}$

Compute the values.

  $=>\left(a+5\right)=\pm 5$

Now, equate both values, to get the answers of a.

  $=>a+5=5$

Simplify for a .

  $=>a=0$

This one value.

  $=>a+5=-5$

Computing the other value.

  $=>a=10$