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655-705 (Hard)|   Algebra|      
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Bunuel
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 01 Feb 2021, 07:15
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55% (hard)

Question Stats:

65% (02:02) correct 35% (02:11) wrong based on 288 sessions

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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 = 0 has equal roots?

A. 0 or -1
B. 0 or 1
C. 0 or 2
D. 0 or 3
E. 0 or 4
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D
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 01 Feb 2021, 07:32
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Bunuel
For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 = 0 has equal roots?

A. 0 or -1
B. 0 or 1
C. 0 or 2
D. 0 or 3
E. 0 or 4
Show more

For equal Roots, \(b^2 = 4ac\) if equation is \(ax^2+bx+c=0\)



\((K + 1)x^2 – 2(K – 1)x + 1 = 0\)

\([-2(k-1)]^2 = 4*(k+1)*1\)

i.e. \(4k^2 + 4 - 8k = 4k+4\)

i.e. \(4k^2 = 12k\)

i.e. k = 3 or 0

Answer: Option D
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 09 Feb 2021, 07:27
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Bunuel Could you please explain how is the correct answer (A)? Since I think when k= -1, it doesn't remain a quadratic equation.

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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 09 Feb 2021, 07:39
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Bunuel Could you please explain how is the correct answer (A)? Since I think when k= -1, it doesn't remain a quadratic equation.

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OA is D. Edited. Thank you!
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 09 Feb 2021, 09:21
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Asked: For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 = 0 has equal roots?


4(K-1)^2 - 4(K+1) = 0
K^2 - 2K + 1 - K - 1 = 0
K^2 - 3K = 0

K = 0 or 3

IMO D
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 12 Feb 2021, 11:18
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Bunuel
For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 = 0 has equal roots?

A. 0 or -1
B. 0 or 1
C. 0 or 2
D. 0 or 3
E. 0 or 4

For equal Roots, \(b^2 = 4ac\) if equation is \(ax^2+bx+c=0\)



\((K + 1)x^2 – 2(K – 1)x + 1 = 0\)

\([-2(k-1)]^2 = 4*(k+1)*1\)

i.e. \(4k^2 + 4 - 8k = 4k+4\)

i.e. \(4k^2 = 12k\)

i.e. k = 3 or 0

Answer: Option D
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Can you please elaborate a little further on your answer? I don't see how we end up with \([-2(k-1)]^2 = 4*(k+1)*1\)...? is this a shortcut?
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 06 Mar 2021, 01:13
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erv20
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Bunuel
For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 = 0 has equal roots?

A. 0 or -1
B. 0 or 1
C. 0 or 2
D. 0 or 3
E. 0 or 4

For equal Roots, \(b^2 = 4ac\) if equation is \(ax^2+bx+c=0\)



\((K + 1)x^2 – 2(K – 1)x + 1 = 0\)

\([-2(k-1)]^2 = 4*(k+1)*1\)

i.e. \(4k^2 + 4 - 8k = 4k+4\)

i.e. \(4k^2 = 12k\)

i.e. k = 3 or 0

Answer: Option D


Can you please elaborate a little further on your answer? I don't see how we end up with \([-2(k-1)]^2 = 4*(k+1)*1\)...? is this a shortcut?
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erv20

I have highlighted in my explanation that \(b^2 = 4ac\) must eb satisfied for real and EQUAL roots of a quadratic equation, which is in form \(ax^2+bx+c=0\)

Compare \((K + 1)x^2 – 2(K – 1)x + 1 = 0\) with \(ax^2+bx+c=0\)

a = (K + 1), b = - 2(K – 1) and c = 1

I hope this help! :)
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For what values of K the quadratic equation (K + 1)x^2 – 2(K – 1)x + 1 19 Feb 2025, 15:17
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