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13 Mar 2020, 01:58
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A
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Difficulty:
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Question Stats:
52% (01:57) correct
48%
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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
Are You Up For the Challenge: 700 Level Questions
A. 0, -1
B. -1, 1
C. 0, 1
D. -1, 2
E. 2, 3
Are You Up For the Challenge: 700 Level Questions
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)
Posted from my mobile device
\(x^{2} +px +(1 —p) = 0\)
—> \(x_1 = 1 —p\)
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)
Posted from my mobile device
13 Mar 2020, 06:21
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Bunuel
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
