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10 Feb 2023, 02:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
57% (03:16) correct
43%
(02:47)
wrong
based on 181
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What is the equation where roots are \((p + mq)\) and \((q + mp)\), if p and q are roots of \(2x^2 – 5x + 2 = 0\)?
A. \(4x^2 - 10(m + 1)x - 4m^2 - 17m - 4 = 0\)
B. \(4x^2 - 10(m + 1)x - 4m^2 + 17m + 4 = 0\)
C. \(4x^2 - 10(m + 1)x - 4m^2 - 17m + 4 = 0\)
D. \(4x^2 - 10(m + 1)x + 4m^2 - 17m - 4 = 0\)
E. \(4x^2 - 10(m + 1)x + 4m^2 + 17m + 4 = 0\)
700-Level question!
A. \(4x^2 - 10(m + 1)x - 4m^2 - 17m - 4 = 0\)
B. \(4x^2 - 10(m + 1)x - 4m^2 + 17m + 4 = 0\)
C. \(4x^2 - 10(m + 1)x - 4m^2 - 17m + 4 = 0\)
D. \(4x^2 - 10(m + 1)x + 4m^2 - 17m - 4 = 0\)
E. \(4x^2 - 10(m + 1)x + 4m^2 + 17m + 4 = 0\)
700-Level question!
10 Feb 2023, 04:12
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p and q are roots of \(2x^2–5x+2=0\)
the equation can be rewritten as \(x^2 -\frac{5}{2}x + 1 = 0\\
x^2 - 2x - \frac{1}{2}x +1 = 0\\
(x-2)(x-\frac{1}{2}) = 0\)
therefore the two roots are x = 2, \(\frac{1}{2 }\)
But what value should be assigned to p and q ?
Now to determine the equation with roots (p + mq) and (q + mp)
On careful observation we find that either of the value (2 or \(\frac{1}{2 }\)) can be assigned to p or q. In both cases the roots (p + mq) and (q + mp) will remain (2 + \(\frac{m}{2}\)) and (\(\frac{1}{2 }\)+ 2m).
Therefore,
sum of roots = \(\frac{5}{2 }+ \frac{5m}{2}\)
product of roots = \((2 + \frac{m}{2})*(\frac{1}{2} + 2m) = \frac{(4 + 17m + 4m^2)}{4 }\)
Now the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
\(x^2 - (\frac{5}{2} + \frac{5m}{2})x + \frac{(4 + 17m + 4m^2)}{4} = 0\\
4x^2−10(m+1)x+4m^2+17m+4 = 0\)
Answer E
the equation can be rewritten as \(x^2 -\frac{5}{2}x + 1 = 0\\
x^2 - 2x - \frac{1}{2}x +1 = 0\\
(x-2)(x-\frac{1}{2}) = 0\)
therefore the two roots are x = 2, \(\frac{1}{2 }\)
But what value should be assigned to p and q ?
Now to determine the equation with roots (p + mq) and (q + mp)
On careful observation we find that either of the value (2 or \(\frac{1}{2 }\)) can be assigned to p or q. In both cases the roots (p + mq) and (q + mp) will remain (2 + \(\frac{m}{2}\)) and (\(\frac{1}{2 }\)+ 2m).
Therefore,
sum of roots = \(\frac{5}{2 }+ \frac{5m}{2}\)
product of roots = \((2 + \frac{m}{2})*(\frac{1}{2} + 2m) = \frac{(4 + 17m + 4m^2)}{4 }\)
Now the desired equation is -
\(x^2\) - (sum of roots)\(x\) + product of roots = 0
\(x^2 - (\frac{5}{2} + \frac{5m}{2})x + \frac{(4 + 17m + 4m^2)}{4} = 0\\
4x^2−10(m+1)x+4m^2+17m+4 = 0\)
Answer E
01 Jul 2024, 01:00
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I wasted 5 minutes in this question. it could have been easily solved by setting m=0, and cheeky the options …
Posted from my mobile device
Posted from my mobile device
