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08 Nov 2021, 10:57
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If a is a constant, and \(x \)= \(−2\) is one solution to the equation \(x^3 + (2a−1)x^2 + ax\) = \(12\) then which of the following is another solution to this equation?
A) \(x \)= −6
B) \(x\) = −1
C) \( x\) = 2
D) \(x\) = 3
E) \(x \) = 4
A) \(x \)= −6
B) \(x\) = −1
C) \( x\) = 2
D) \(x\) = 3
E) \(x \) = 4
08 Nov 2021, 18:53
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nislam
If a is a constant, and \(x \)= \(−2\) is one solution to the equation \(x^3 + (2a−1)x^2 + ax\) = \(12\) then which of the following is another solution to this equation?
A) \(x \)= −6
B) \(x\) = −1
C) \( x\) = 2
D) \(x\) = 3
E) \(x \) = 4
A) \(x \)= −6
B) \(x\) = −1
C) \( x\) = 2
D) \(x\) = 3
E) \(x \) = 4
Putting x = -2 will satisfy the eqn,
we get a = 4.
By putting a = 4 in eqn, The eqn is reduced to x^3 + 7 x^2 + 4x = 12
x (x^2 + 7x +4) = 12
Checking with option, x= -6 satisfies the eqn
Option A
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12 Jul 2022, 00:45
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Putting x = -2 into equation, we have:
-8 + (2a-1)4 - 2a - 12 = 0
Solve the equation: a = 4
Then, putting a = 4 into the first equation, we have:
x^3+ 7x^2 + 4x - 12 = 0
Because we know that it has x=-2 is one of the solution, therefore, we rewrite the equation:
x^3 + 2x^2 + 5x^2 + 10x - 6x - 12 = 0
=> (x+2) (x^2 + 5x - 6) = 0
We now have all 3 solutions, which are: -2, -6, 1.
Ans: A
-8 + (2a-1)4 - 2a - 12 = 0
Solve the equation: a = 4
Then, putting a = 4 into the first equation, we have:
x^3+ 7x^2 + 4x - 12 = 0
Because we know that it has x=-2 is one of the solution, therefore, we rewrite the equation:
x^3 + 2x^2 + 5x^2 + 10x - 6x - 12 = 0
=> (x+2) (x^2 + 5x - 6) = 0
We now have all 3 solutions, which are: -2, -6, 1.
Ans: A
