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08 Nov 2018, 01:45
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
68% (02:21) correct
32%
(02:41)
wrong
based on 360
sessions
History
Date
Time
Result
Not Attempted Yet
Which of the following are roots of an equation \(10x^{-2}+x^{-1}-21=0\)
\(A. \frac{-2}{3}\) or \(\frac{5}{7}\)
\(B. \frac{-2}{3}\) or \(\frac{7}{5}\)
\(C. \frac{2}{3}\) or \(\frac{-5}{7}\)
\(D. \frac{2}{3}\) or \(\frac{-7}{5}\)
\(E. \frac{3}{2}\) or \(\frac{5}{7}\)
\(A. \frac{-2}{3}\) or \(\frac{5}{7}\)
\(B. \frac{-2}{3}\) or \(\frac{7}{5}\)
\(C. \frac{2}{3}\) or \(\frac{-5}{7}\)
\(D. \frac{2}{3}\) or \(\frac{-7}{5}\)
\(E. \frac{3}{2}\) or \(\frac{5}{7}\)
08 Nov 2018, 02:36
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Attached is the solution:
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General Discussion
08 Nov 2018, 02:29
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MathRevolution
\(\frac{10}{x^2}+\frac{1}{x}-21=0\)
Or, \(10+x-21x^2=0\)
Or, \(-21x^2+x+10=0\) (Standard form of quadratic equation)
Sum of the roots=-(-1/21)=1/21
Product of the roots=-10/21
Only option(A) matches to the above criteria.
Ans. (A)
11 Nov 2018, 23:06
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=>
\(10x^{-2}+x^{-1}-21=0\)
\(=> 10+x-21x^2=0\) by multiplying \(x^2\)
\(=> 21x^2-x-10=0\) by multiplying \((-1)\)
\(=> (3x+2)(7x-5)=0\)
\(=> x =\frac{-2}{3}\) or \(x = \frac{5}{7}\)
Therefore, the answer is A.
Answer: A
\(10x^{-2}+x^{-1}-21=0\)
\(=> 10+x-21x^2=0\) by multiplying \(x^2\)
\(=> 21x^2-x-10=0\) by multiplying \((-1)\)
\(=> (3x+2)(7x-5)=0\)
\(=> x =\frac{-2}{3}\) or \(x = \frac{5}{7}\)
Therefore, the answer is A.
Answer: A
