Question

If the equation x4−(k−1)x2+(2−k)=0 has three distinct real roots, then the possible value(s) of k is/are{2}{√2−1,2}{√5−1}{2√2,√3−√2}

Solution

The correct option is A {2}x4−(k−1)x2+(2−k)=0   ⋯(1) Assuming x2=t, t2−(k−1)t+(2−k)=0   ⋯(2) Let roots of the equation (2) be t1,t2 (t1≤t2) Equation (1) will have three distinct real roots iff for equation (2), t1=0,  t2>0 (i) f(0)=0⇒2−k=0⇒k=2   ⋯(3)(ii) −b2a>0⇒k−12>0⇒k>1   ⋯(4) From equation (3) and (4), k=2

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