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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.

 

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 Apr 25, 2020
 #1
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Let's first expand our LHS(left hand side).

We get:

\(2x^2-3x-35 = -43 + jx\)

Adding 43 on both sides,

\(2x^2-3x+8 = jx\)

Subtracting jx on both sides, 

\(2x^2-(3+j)x + 8 = 0\)

Next, realize that if we have exactly one real solution to this quadratic, that implies it must be a double root; in other words, two of the same real root(otherwise if it was one real and one imaginary, the quadratic wouldn't have real coefficients). In order for this to be true, the discriminant of our quadratic must equal 0.

Generally for quadratics, the discriminant is:

\(\sqrt{b^2-4ac}\)

In this case, our discriminant for this quadratic would be:

\(\sqrt{(3+j)^2-4*8*2} = 0\)

\(\)Squaring both sides, we get:
\((3+j)^2 - 4*8*2 =0\)

\(9+6j+j^2-4*8*2 = 0\)

\(9+6j+j^2-64 = 0\)

\(j^2+6j-55 = 0\)

Factoring this out, we get:
\((j+11)(j-5) = 0\)

Our two values for j that satisfy this equation are then -11 and 5

 Apr 25, 2020

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