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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.

 Dec 20, 2019
 #1
avatar+119 
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We can start by expanding the equation (2x + 7)(x - 5) = -43 + jx ---> 2x^2 - (j+3)x + 8 = 0.

Now we know that when the discriminant (b^2 - 4ac) is equal to 0, we have one real solution.

So we plug in our values and we get j^2 + 6j - 55 = 0.

Solving this gives us j = -11, 5

 Dec 20, 2019
 #2
avatar+11146 
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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.

laugh

 Dec 20, 2019

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