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if the roots of the equation 1/x+a + 1/x+b =1/c are equql in magitude but opposite in signs , then their product is (a)- ½ (a squqre+b square) (b)- -1/2 ( a square+ b square) (c)- 1/2ab (d)- -1/2ab

if the roots of the equation 1/x+a   +  1/x+b   =1/c  are equql in magitude but opposite in signs , then their product is   (a)-  ½ (a squqre+b square)  (b)-  -1/2 ( a square+ b square)  (c)-  1/2ab  (d)-  -1/2ab 

Grade:12th pass

2 Answers

noogler
489 Points
8 years ago
aren’t the first and second optins are same?
whatever it may be the correct ans is -1/2(a2+b2)
 
on solving the given equation into the form ax2+bx+c=0
 
we get x2+x(a+b+2c)+ab-(a+b)c=0 
given the two roots of it are equal in magnitude but opp in sign
i.e,  sum of roots =0
      from given eq. sum of roots=-b/a=-(a+b+2c)
                 -(a+b+2c)=0............................a+b/2=c
substitute c=a+b/2 in product of roots = c/a =ab-(a+b)c
on solving it u will get the ans :-)
 
Rishi Sharma
askIITians Faculty 646 Points
3 years ago
Dear Student,
Please find below the solution to your problem.

solving the given equation into the form ax2+bx+c=0
we get x2+x(a+b+2c)+ab-(a+b)c=0
given the two roots of it are equal in magnitude but opp in sign
i.e, sum of roots =0
from given eq. sum of roots=-b/a=-(a+b+2c)
-(a+b+2c)=0............................a+b/2=c
substitute c=a+b/2 in product of roots = c/a =ab-(a+b)c
on solving it u will get the answer

Thanks and Regards

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