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If t > 0, and s is the only value of x for which x2 + tx + 100 = 0, th
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27 Dec 2021, 12:57
27 Dec 2021, 12:57
Expert Reply
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Question Stats:
62% (01:46) correct
37% (02:04) wrong based on 16 sessions
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If t > 0, and s is the only value of x for which \(x^2 + tx + 100 = 0\), then s + t =
(A) -20
(B) -10
(C) 0
(D) 10
(E) 20
(A) -20
(B) -10
(C) 0
(D) 10
(E) 20
If t > 0, and s is the only value of x for which x2 + tx + 100 = 0, th
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29 Dec 2021, 19:43
29 Dec 2021, 19:43
1
\(x^2 + tx + 100 = 0\)
\(x^2 + 100\) will always be greater than \(0\) so \(tx\) will be -ve
we are given that \(t>0\) so \(x<0\)
\(x^2 + tx + 100 = 0\)
\(x^2 - t(-x) + 100 = 0\)
This is similar with \((x+10)^2 =\) \(x^2 + 20x + 100 = 0\)
\(tx = 20x\)
\(t = 20\)
only possible value for \(x\) is \(-10\)
\(s = x = -10\)
\(s+t = -10+20 = 10\)
Answer D
\(x^2 + 100\) will always be greater than \(0\) so \(tx\) will be -ve
we are given that \(t>0\) so \(x<0\)
\(x^2 + tx + 100 = 0\)
\(x^2 - t(-x) + 100 = 0\)
This is similar with \((x+10)^2 =\) \(x^2 + 20x + 100 = 0\)
\(tx = 20x\)
\(t = 20\)
only possible value for \(x\) is \(-10\)
\(s = x = -10\)
\(s+t = -10+20 = 10\)
Answer D
Re: If t > 0, and s is the only value of x for which x2 + tx + 100 = 0, th
[#permalink]
01 Sep 2022, 23:16
01 Sep 2022, 23:16
1
Why E is not correct option?
For me option D and E both are correct. As it both satisfy the condition t>0.
For me option D and E both are correct. As it both satisfy the condition t>0.
