w What is twice the nonnegative difference between the solutions of the equation $x^2-7x-4=40$?
First we need to find the solutions of the equation...
x2 - 7x - 4 = 40
Subtract 40 from both sides so that the right side is zero.
x2 - 7x - 44 = 0
Now we can factor the left side like this..
(x + 4)(x - 11) = 0
Set each factor equal to zero and solve each for x .
x + 4 = 0 or x - 11 = 0
x = -4 or x = 11
The solutions of the equation are -4 and 11 .
Twice the nonnegative difference of -4 and 11 is...
2 * | -4 - 11 | = 2 * | -15 | = 2 * 15 = 30
First we need to find the solutions of the equation...
x2 - 7x - 4 = 40
Subtract 40 from both sides so that the right side is zero.
x2 - 7x - 44 = 0
Now we can factor the left side like this..
(x + 4)(x - 11) = 0
Set each factor equal to zero and solve each for x .
x + 4 = 0 or x - 11 = 0
x = -4 or x = 11
The solutions of the equation are -4 and 11 .
Twice the nonnegative difference of -4 and 11 is...
2 * | -4 - 11 | = 2 * | -15 | = 2 * 15 = 30