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Bunuel
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 13 Mar 2020, 01:58
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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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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 13 Mar 2020, 02:38
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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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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 13 Mar 2020, 06:21
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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
Show more

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)[(1 - p) + p + 1] = 0
Simplify: (1 - p)[2] = 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
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 03 Jun 2020, 11:17
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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.
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 10 Jun 2020, 02:54
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sambitspm
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.
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Why is the sum and the product as stated?
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 11 Jun 2020, 10:38
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binzdelabinz
sambitspm
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.
Why is the sum and the product as stated?
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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.
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 11 Jun 2024, 05:50
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BrentGMATPrepNow

Bunuel
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

 
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)[(1 - p) + p + 1] = 0
Simplify: (1 - p)[2] = 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
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­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
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 11 Jun 2024, 06:08
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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
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If (1 p) is a root of quadratic equation x^2 + px + (1 p) = 0 then 16 Aug 2025, 07:41
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