### Instruction: For what values of $j$ does the equation $(2x+7)(x-5) = -43 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.### Response: To solve this problem, I need to find out when the given quadratic equation has a discriminant of zero, since that means it has one repeated root.
To do that, I need to expand the left-hand side and rearrange the equation in standard form, $ax^2 + bx + c = 0$.
Expanding the left-hand side, I get $(2x+7)(x-5) = 2x^2 - 10x + 7x - 35 = 2x^2 - 3x - 35$.
Moving everything to one side, I get $2x^2 - 3x - 35 + 43 - jx = 0$.
Simplifying, I get $2x^2 - (3 + j)x + 8 = 0$.
Now, I can use the formula for the discriminant, $D = b^2 - 4ac$, and set it equal to zero.
Plugging in the values of $a$, $b$, and $c$, I get $(3 + j)^2 - 4(2)(8) = 0$.
Expanding and simplifying, I get $9 + 6j + j^2 - 64 = 0$.
Rearranging, I get $j^2 + 6j - 55 = 0$.
This is a quadratic equation in $j$, so I can use the quadratic formula to find the values of $j$ that make it true.
Using the quadratic formula, I get $j = \frac{-6 \pm \sqrt{6^2 - 4(1)(-55)}}{2(1)} = \frac{-6 \pm \sqrt{256}}{2} = \frac{-6 \pm 16}{2}$.
Simplifying, I get two possible values for $j$: $j = -11$ or $j = 5$.
Therefore, the equation has exactly one real solution when $j = -11$ or $j = 5$.