The quadratic equation above, where c is a constant and x is a variabl

Question

x^2 - 12x + c = 0 
The quadratic equation above, where c is a constant and x is a variable, has two distinct roots, one of which is the square of the other. What is the sum of all possible value of c ?

A. -91
B. -67
C. -64
D. -37
E. 27

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ArunSharma12 · Answer

x2−12x+c=0 
The quadratic equation above, where c is a constant and x is a variable, has two distinct roots, one of which is the square of the other. What is the sum of all possible value of c ?

A. -91
B. -67
C. -64
D. -37
E. 27

let one root be a, then the other root will be  a^2

a + a^2 = 12  (sum of the roots =  \frac{-b}{a} )
a = -4, 3

a^3 = c  (product of the roots =  \frac{c }{a} )
c can be -64, 27. Total sum = -37
Ans: D

GMATWhizTeam · Answer

Solution

• Let the root of the given quadratic equation be  p  and  p^2. 
• So, Product of the roots  = p*p^2 = \frac{c}{1} ⟹ c = p^3 …………. Eq. (i) 
• And, sum of the roots  = p + p^2 = \frac{-(-12)}{1} = 12 ⟹ p^2 + p -12 = 0 
 o Or,  (p + 4) (p - 3) = 0 ⟹ p = -4  or  3   
• Substituting the above obtained value of p in Eq.(i), we get two possible values of c .i.e. 
 o  c = (-4) ^3 = -64 
o Or,  c = 3^3 = 27   
• Hence, the required sum  = -64 + 27 = -37  
Thus, the correct answer is  Option D.

CrackverbalGMAT · Answer

In this question, although it looks like we have to deal with the quadratic equation  x^2 -12x + c = 0, we will actually have to solve another quadratic equation which is formed in terms of the roots of the former.

Let α and β be the roots of the equation   x^2  – 12x + c = 0. Comparing with  the standard form of a quadratic equation , a x^2  + bx+c = 0, we have,
 a=1, b = -12 and c=c.

The sum of roots of a quadratic equation is given by –(b/a) and the product of the roots is given by c/a.

The question says that one of the roots of the equation is the square of the other. Let β =  α^2 . Then,

Product of roots = c / 1 =  α^3  and Sum of roots = -(-12/1) = α +  α^2 , which can be simplified as,
 α^2  + α – 12 = 0. Now, THIS is the other quadratic equation I was talking about. Solving this will give us the possible values for α and hence the possible values for c.

Factorising the equation above, we have, 
 α^2  + 4α - 3α -12 = 0 (since the factors of -12 are 4 and -3). Solving this equation by taking common terms, we have α = -4 or α = 3. Therefore, there are two possible values for α^3 i.e. -64 and 27. The sum of these two values is -37.

The correct answer option is D.

Hope that helps!

lacktutor · Answer

The equation has two roots:  (x -x_1)*(x -x_2)=0 
let  x_1  be  a 
-->  x_2  must be  a^{2} 
 (x -a)*(x -a^{2})=0

x^{2} -x(a^{2} + a) +a^{3}=0 
 x^{2}−12x+c=0

-->  a^{2} + a= 12 ,  a^{3}=c  
 (a -3)(a +4) =0

-->  a= 3  and  a= -4  
-->  c  could be  27  and  -64  
 27- 64 = -37

The answer is D.

GMATinsight · Answer

x^2 - 12x + c = 0

Sum of the roots, p+q = -(-12)/1 = 12
Product of teh roots, p*q = c/1 = c

Also, p = q^2

i.e. q^3 = c

Also, p+q = q^2+q = q(q+1) = 12

i.e. q = 3 or q = -4

c = q^3 = 3^3 or (-4)^3 = 27 or -64

Sum = 27-64 = -37

Answer: Option D

ScottTargetTestPrep · Answer

If we let r and s be the two distinct roots of the equation, we have:

(x - r)(x - s) = 0

x^2 - rx - sx + rs = 0

x^2 - (r + s)x + rs = 0

Therefore, we see that r + s must be 12 and rs must be c. Since one root is the square of the other, we can let s = r^2. Thus we have:

r + r^2 = 12

r^2 + r - 12 = 0

(r - 3)(r + 4) = 0

r = 3 or r = -4

If r = 3, then s = r^2 = 9 (notice that r + s = 12) and c = rs = 27. If r = -4, then s = r^2 = 16 (notice that again r + s = 12) and c = rs = -64. Therefore, the sum of all possible values of c is 27 + (-64) = -37.

Answer: D

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