If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then

Question

If  (1 − p)  is a root of quadratic equation  x^2 + px + (1 − p) = 0  then its roots are

A. 0, -1
B. -1, 1
C. 0, 1
D. -1, 2
E. 2, 3

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lacktutor · Answer

Given that: 
 x^{2} +px +(1 —p) = 0 
—>  x_1 = 1 —p 
Well, 
—>  (1–p)^{2} + p(1–p) + (1–p) = 0

(1 —p)( 1–p +p +1 ) = 0

2( 1–p) = 0 
—>  p= 1

—>  x^{2} + x +0 =0

x( x+1) = 0

x = —1 ,  x= 0

Answer(A)

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BrentGMATPrepNow · Answer

If (1 − p) is a  root , then  x = (1 − p)  is a  solution  to the equation x² + px + (1 − p) = 0 
Replace x with  (1 - p)  to get:  (1 - p) ² + p (1 - p)  + (1 - p) = 0
Factor out the (1 - p) to get: (1 - p)  = 0
Simplify: (1 - p)  = 0
So, p =  1

If p =  1 , our equation, x² + px + (1 − p) = 0, becomes x² + ( 1 )x + (1 −  1 ) = 0
Simplify: x² + x = 0
Factor: x(x + 1) = 0
So, EITHER x = 0 OR x = -1

Answer: A

Cheers,
Brent

sambitspm · Answer

One can do this by substituting (1-p) in the equation. Hell, I also thought of it. But then I thought of playing smart. And here is my approach.

binzdelabinz · Answer

Why is the sum and the product as stated?

sambitspm · Answer

Hi  binzdelabinz , for a quadratic equation x^2+ax+b = 0, the sum of the roots is -a and product is b. So I used that concept to solve the problem.

RC_xx · Answer

I got the correct solution but with different approach and I was wondering if my method could work for any other problem or I got lucky.

If 1-p is the root then 1-p = 0, hence p =1, then sostitute P= 1 and with semplification give me x^2+x=0 then x(x+1)= 0, x= 0 and x=-1

Kinshook · Answer

Asked: If  (1 − p)  is a root of quadratic equation  x^2 + px + (1 − p) = 0  then its roots are

Since (1-p) is  a root of quadratic equation  x^2 + px + (1 − p) = 0 , it must satisfy the equation.

(1-p)^2 + p(1-p) + (1-p) = 0
(1-p) { (1-p) + p + 1 } = 0
2(1-p) = 0

p = 1
1-p = 0

Let the other root be x.

Sum of roots = - p = x; x = -p = - 1;

Roots = {0,-1}

IMO A

SKP220292 · Answer

Don't know why I did not replace value of x as 1-p into the equation, probably would have been faster. But either way following is an alternative (but probably indirect) approach in case it helps anyone.

My approach was that product of roots here is 1-p. Meaning that the other root can be 1 or 0. Putting x=1 in the equation gives us 2 and not zero, so 1 cannot be a root. Which means the other root has to be 0. Which means product of roots = 1-p = 0. Hence, p=1.

Putting P=1 in the equation, we get roots 0,-1

If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then

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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then

Prep Toolkit

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