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Question
Given that p and q are the roots of the equation x2 – ax +b =0 and Dn= pn+qn. Find the value of Dn+1
Given that p and q are the roots of the equation x2 – ax +b =0 and Dn= pn+qn. Find the value of Dn+1
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Solution
The correct option is A aDn-bDn-1
Option (a)
Best way to proceed is by assumption of values.
Assume a quadratic equation. Let’s take x2+5x-6=0. Here the roots are p=1 and q=-6
Thus, a = sum of roots = -5 and b= product of roots = -6
D1= p1+q1 = -5
D0 = 2
D2 = 37
Assume n=1
We need to find Dn+1 = D2 = 37
Look in the answer options for 37
Option a is the only one which gives (-5)(-5)-(-6)(2) = 37
Option (a)
Best way to proceed is by assumption of values.
Assume a quadratic equation. Let’s take x2+5x-6=0. Here the roots are p=1 and q=-6
Thus, a = sum of roots = -5 and b= product of roots = -6
D1= p1+q1 = -5
D0 = 2
D2 = 37
Assume n=1
We need to find Dn+1 = D2 = 37
Look in the answer options for 37
Option a is the only one which gives (-5)(-5)-(-6)(2) = 37
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