Q. 204.7( 12 Votes )

If a and b are real and ab then show that the roots of the equation (a – b)x2 + 5(a + b)x – 2(a – b) = 0 are real and unequal.

Answer :

Comparing with standard quadratic equation ax2 + bx + c = 0

a = (a – b) b = 5(a + b) c = – 2(a – b)


Discriminant D = b2 – 4ac


= [5(a + b)]2 – 4(a – b)(– 2(a – b))


= 25(a + b)2 + 8(a – b)2


Since a and b are real and ab then (a + b)2 > 0 (a – b)2 > 0


8(a – b)2 > 0 – – – – (1) product of two positive numbers is always positive


25(a + b)2 > 0 – – – – (2) product of two positive numbers is always positive


Adding (1) and (2) we get


8(a – b)2 + 25(a + b)2 > 0 (sum of two positive numbers is always positive)


D > 0


Hence the roots of given equation are real and unequal.


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