### 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: I notice that the equation is quadratic in $x$, so it has exactly one real solution if and only if its discriminant is zero.
That is, the part under the square root in the quadratic formula is zero.
So I want to find the discriminant of this equation and set it equal to zero.
To do that, I need to write the equation in the standard form $ax^2 + bx + c = 0$.
I can expand the left-hand side using the distributive property: $(2x+7)(x-5) = 2x^2 - 10x + 7x - 35 = 2x^2 - 3x - 35$.
Then I can move everything to one side and simplify: $2x^2 - 3x - 35 + 43 - jx = 0$, which gives me $2x^2 - (3+j)x + 8 = 0$.
Now I can identify the coefficients of the quadratic: $a = 2$, $b = -(3+j)$, and $c = 8$.
The discriminant is given by $b^2 - 4ac$, so I want to plug in these values and set it equal to zero: $(-(3+j))^2 - 4(2)(8) = 0$.
I can simplify this expression by squaring the binomial and expanding the product: $(9 + 6j + j^2) - 64 = 0$.
Then I can rearrange the terms and factor the quadratic: $j^2 + 6j - 55 = 0$, which gives me $(j+11)(j-5) = 0$.
Therefore, the discriminant is zero if and only if $j = -11$ or $j = 5$.