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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
avatar+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

 

Answer: -5.89, 3.89

 Jan 13, 2024
 #2
avatar+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)   

 

 

cool cool cool

 Jan 14, 2024

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