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# If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h

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Joined: 27 Oct 2017
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WE: Business Development (Energy and Utilities)
If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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27 Oct 2018, 09:27
00:00

Difficulty:

45% (medium)

Question Stats:

54% (01:11) correct 46% (01:48) wrong based on 23 sessions

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If a and b are constants, and the equation $$x^3 +ax^2 +bx = 64$$ has precisely one solution for x, what is the value of b?
A) 48
B) 16
C) 24
D) -16
E) -67

Weekly Quant Quiz #6 Question No 9

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Senior DS Moderator
Joined: 27 Oct 2017
Posts: 1066
Location: India
GPA: 3.64
WE: Business Development (Energy and Utilities)
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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27 Oct 2018, 09:27
Because if this cubic polynomial has only one root, 4 , then the equation can be written as:

(x - 4)^3 = 0

And then when we expand this we will get :

$$x^3 - 3*4*x^2 + 3*4^2*x - 64 = 0$$

Hence comparing to equation given, we get b = 48
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Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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27 Oct 2018, 22:47
x=4. using sum of multiplication of roots b=48
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Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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27 Oct 2018, 23:31
$$x^3+ax^2+bx-64 = 0$$
let the root = k
product of roots = -(-64)/1 = 64
since the roots are the same , $$k^3 = 64 ; k = 4$$

sum of root = -a/1
a = -12

putting the value of x in the equation ,
64 + 16a + 4b = 64
16a+4b = 0
4a + b = 0
b = -4a = 48
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Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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28 Oct 2018, 02:47
I got A for the answer.

x3+ax2+bx=64
<=> x^3 + ax^2 + bx - 64 = 0

Vieta's theorem
x1*x2*x3 = -(-64)/1

Since there is only 1 value of x (x1, x2 and x3 are the same), meaning x^3 = 64 => x=4

x1x2 +x2x3 + x3x1 = b/1
or 4*4 + 4*4 +4 *4 = 16 + 16 + 16 = 48
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h &nbs [#permalink] 28 Oct 2018, 02:47
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