+624 For what values of $j$ does the equation $(2x + 7)(x - 4) = -31 + jx$ have exactly one real solution? Express your answer as a list of numbers, separated by commas.
+289 For there to be 1 solution, the discriminant must be equal to 0, so b^2-4ac = 0
Simplify:
\(2x^2-x-jx+3=0\)
a = 2, b = -1, and c = 3
Pluggng a and c in, we get:
\(\sqrt{b^2-24}=0\)
This means that \(b^2 = 24\).
\(b=\pm \sqrt 24\)
\(b=\pm4.89\)
We already have -1, and we are subtracting j, so we have:
-1 - j = 4.89
-j = 5.89
j = -5.89
and,
-1 - j = -4.89
-j = -3.89
j = 3.89
the values of j are -5.89 and 3.89
Answer: -5.89, 3.89
+128663 (2x + 7) (x - 4) = -31 + jx
2x^2 - x - 28 = -31 + jx
2x^2 - (1 + j) + 3 = 0
To have only one solution
(1 + j)^2 - 4(2)(3) = 0
(1 + j)^2 - 24 = 0
(1 + j)^2 = 24 take both roots
j + 1 = sqrt (24) j + 1 = -sqrt (24)
j + 1 = 2sqrt (6) j + 1 = -2sqrt (6)
j = -1 + 2sqrt (6) j = -1 - 2sqrt (6)
