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If (1 − p) is a root of quadratic equation x^2 + px + (1 − p
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07 Apr 2020, 03:34
07 Apr 2020, 03:34
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41% (02:03) wrong based on 41 sessions
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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
Kudos for the right answer and explanation
A. 0, -1
B. -1, 1
C. 0, 1
D. -1, 2
E. 2, 3
Kudos for the right answer and explanation
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If (1 − p) is a root of quadratic equation x^2 + px + (1 − p
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07 Apr 2020, 05:13
07 Apr 2020, 05:13
4
Carcass wrote:
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
Kudos for the right answer and explanation
A. 0, -1
B. -1, 1
C. 0, 1
D. -1, 2
E. 2, 3
Kudos for the right answer and explanation
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
Re: If (1 p) is a root of quadratic equation x^2 + px + (1 p
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16 Feb 2023, 03:06
16 Feb 2023, 03:06
Hi Brent GreenlightTestPrep, I multiply 2 into (1-p) and got p=2, what's wrong here?
Simplify: (1 - p)[2] = 0
So, p = 1
Simplify: (1 - p)[2] = 0
So, p = 1
