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 a
b 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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