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#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h

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Retired Moderator V
Joined: 27 Oct 2017
Posts: 1272
Location: India
GPA: 3.64
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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1 00:00

Difficulty:   75% (hard)

Question Stats: 29% (02:02) correct 71% (02:01) wrong based on 51 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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Retired Moderator V
Joined: 27 Oct 2017
Posts: 1272
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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1
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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Intern  B
Joined: 26 Mar 2018
Posts: 1
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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x=4. using sum of multiplication of roots b=48
Manager  G
Joined: 14 Jun 2018
Posts: 212
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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$$x^3+ax^2+bx-64 = 0$$
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
Intern  B
Joined: 21 Aug 2018
Posts: 3
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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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
Senior Manager  S
Joined: 12 Sep 2017
Posts: 306
Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  [#permalink]

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gmatbusters wrote:
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

Hello Math experts!

Could someone please explain to me how to use the Viete's theorem in this kind of equation?

Which is a and which is b?

$$x^3 +ax^2 +bx = 64$$

Kind regards! Re: If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h   [#permalink] 17 Jan 2019, 19:44
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# If a and b are constants, and the equation [m]x^3 +ax^2 +bx = 64[/m] h  