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. thank you!

Quadratic equations are the polynomial equations of degree 2 in one variable of the type [tex]f(x) = ax^{2} + bx + c = 0[/tex] where a, b, c, ∈ R and a ≠ 0.
The standard form of a quadratic equation is [tex]ax^{2} + bx + c = 0[/tex] .

Here, the given equation is (2x+7)(x-5) = - 43 + jx

On simplifying the equation we get,

[tex]2x^{2} -10x+7x-35=-43+jx\\2x^{2} -3x-35+43-jx=0\\2x^{2} -3x-jx+8=0\\2x^{2} -(3+j)x+8=0.....(1)\\[/tex]

By comparing the given equation with the standard form of the quadratic equation we get,

a = 2

b = - (3 + j)

c = 8

Therefore,

[tex](-(3+j))^{2} -4\times2\times8=0\\(3+j)^{2} -64=0\\(3+j)^{2} =64\\[/tex]

Take the square root of both sides,

[tex]\sqrt{(3+j)^{2} }=\sqrt{64} \\[/tex]

3 + j = ±8

Therefore,

3 + j = 8

j = 8 - 3

j = 5

or

3 + j = -8

j = -8 - 3

j = -11

Hence, the possible values of 'j' are -11,5.

Learn more about the quadratic equation at https://brainly.com/question/8649555

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