For what values of j does the equation \((2x+7)(x-4)=-30+jx\) have exactly one real solution? Express your answer as a list of numbers, separated by commas.

Iamrockingtoday Jul 4, 2024

#1**0 **

First, we expand the left-hand side of the equation: (2x + 7)(x - 4) = 2x^2 - 8x + 7x - 28 = 2x^2 - x - 28

Then, we rewrite the equation with all terms on one side: 2x^ 2- x - 28 = -30 + jx

Simplifying, we obtain: 2x^2 + (j - 1) x + 2 = 0

For the quadratic equation to have exactly one real solution, the discriminant must be zero. That is, b^2 - 4ac = 0

Substituting a = 2, b = j - 1, c = 2 into the discriminant formula, we get:

(j - 1)^2 - 4(2)(2) = 0

Solving for j, we find:

j^2 - 2j + 1 - 16 = 0

j^2 - 2j - 15 = 0

Factoring, we get:

(j - 5)(j + 3) = 0

Thus, the possible values of j are 5 and -3.

bader Jul 4, 2024