For a∈Q−{1,−2}, the roots of the equation (a2+a−2)x2+(2a2+a−3)x+a2−1=0 are
A
irrational and equal
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B
irrational and distinct
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C
rational and equal
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D
rational and distinct
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Solution
The correct option is D rational and distinct (a2+a−2)x2+(2a2+a−3)x+a2−1=0 Δ=(2a2+a−3)2−4(a2+a−2)(a2−1) =(2a+3)2(a−1)2−4(a−1)(a+2)(a2−1) =(a−1)2[4a2+9+12a−4(a2+3a+2)] =(a−1)2>0asa≠1
So, Δ is a perfect square and positive. Hence, roots are rational and distinct.