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

Question

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

GMATBusters · Answer

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

smriti29 · Answer

x=4. using sum of multiplication of roots b=48

pandeyashwin · Answer

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

Emma263 · Answer

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

jfranciscocuencag · Answer

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!

Kinshook · Answer

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?

x^3 +ax^2 +bx = 64 
 x^3 + ax^2 + bx - 64 = 0

Let x = e; Only solution of the equation

x^3 + ax^2 + bx - 64 = (x-e)^3 = x^3 - 3ex^2 + 3e^2x - e^3= 0

e^3 = 64
e = 4

b = 3e^2 = 3*4^3 = 48

IMO A

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