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27 May 2021, 06:28
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
54% (02:55) correct
46%
(02:45)
wrong
based on 120
sessions
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Not Attempted Yet
If k is a constant, and \( x = −5\) is a solution to the equation \(x^3 + kx^2 − 97x = 210\), which of the following is another solution to this equation?
A) −14
B) −3
C) 5
D) 9
E) 15
A) −14
B) −3
C) 5
D) 9
E) 15
27 May 2021, 07:11
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sjuniv32
As we know one value of x, let us substitute that value to get the value of k.
\(x^3 + kx^2 − 97x = 210\)
\((-5)^3 + k(-5)^2 − 97(-5) = 210\)
\(-125+25k+485=210\)
\(25k=-150\)
\(k=-6\)
So the equation becomes
\(x^3 - 6x^2 − 97x = 210\)
Now, we can substitute the options to see which value fits in if we are not able to factorize it further.
\((-3)^3 + 6*(-3)^2 − 97*-3 = 210\) gives \(210=210\)
Thus the answer is B
We can otherwise factorize it and x will have three values.
\(x^3 - 6x^2 - 97x = 210\)
\(x^3 - 6x^2 - 97x - 210=0\)
Since we know x+5 is a factor, let us get x+5 out of the expression.
\(x^3 +5x^2-5x^2- 6x^2 - 97x - 210=0\)
\(x^2(x +5)-11x^2 - 97x - 210=0\)
\(x^2(x +5)-11x^2 -55x+55x- 97x - 210=0\)
\(x^2(x +5)-11x(x+5)- 42x - 210=0\)
\(x^2(x +5)-11x(x+5)- 42(x +5)=0\)
\((x+5)(x^2-11x-42)=0\)
\((x+5)(x^2-14x+3x-42)=0\)
\((x+5)(x(x-14)+3(x-14))=0\)
\((x+5)(x+3)(x-14)=0\)
So x can be -5, -3 or 14.
B
28 May 2021, 01:47
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sjuniv32
Given:
x^3 + kx^2 − 97x = 210
&
one root: -5
Solving for K:
-125 +25K +97*5 = 210
-125 +25K +485 =210
25K = 335-485
25K = -150
k= -6
Therefore equation becomes:
x^3 - 6x^2 − 97x = 210
Checking with option B as most possibly the other route is negative
x= -3
x^3 - 6x^2 − 97x = 210
-27 - 6*9 - 97*-3 = 210
-27 - 54 + 291 = 210
-81 + 291 = 210
210 = 210
Therefore option B is right answer
