+0

+1
10
2
+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.

Jan 13, 2024

#1
+289
+1

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

Jan 13, 2024
#2
+128732
+1

(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)

Jan 14, 2024