Yahoo Answers: Answers and Comments for Prove that no real value of k exists for the equation x^2kx+2k10=0 has equal roots.(has onlyone value for x)? [Mathematics]
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From Anonymous
enCA
Sun, 03 Apr 2011 18:33:04 +0000
3
Yahoo Answers: Answers and Comments for Prove that no real value of k exists for the equation x^2kx+2k10=0 has equal roots.(has onlyone value for x)? [Mathematics]
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From Hemant: QE : 1x² + (k)x + (2k10) = 0
∴ a = 1, b ...
https://ca.answers.yahoo.com/question/index?qid=20110403183304AA8zLyn
https://ca.answers.yahoo.com/question/index?qid=20110403183304AA8zLyn
Mon, 04 Apr 2011 05:45:17 +0000
QE : 1x² + (k)x + (2k10) = 0
∴ a = 1, b = k, c = 2k  10 .......... (1)
For equal roots, discriminant Δ = b²  4ac = 0.
∴ from (1),
... (k)²  4(1)( 2k  10 ) = 0
∴ 1k²  8k + 20 = 0.
For this kequation, discr. = (8)²  4(1)(20) = 14 < 0
∴ there is NO REAL value of k. ........................................ Q.E.D.
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