Question

If p and q are the roots of QE x2-(alpha-2)x-alpha-1=0 then what is the minimum value of p2+q2

Answer 1

Answer:

Step-by-step explanation:

x2−(a−2)x−a−1=0x2−(a−2)x−a−1=0 = x2−(a−2)x+(−a−1)=0x2−(a−2)x+(−a−1)=0 equivalent to x2−(p+q)x+pq=0x2−(p+q)x+pq=0

Hence p+q=a−2p+q=a−2 & pq=−a−1pq=−a−1

p2+q2p2+q2= (p+q)2−2pq(p+q)2−2pq = (a−2)2−2(−a−1)(a−2)2−2(−a−1) = a2−4a+4+2a+2a2−4a+4+2a+2 = a2−2a+6a2−2a+6

* When a =0 or 2 , p2+q2p2+q2=6 but this is not the min value.

Now , a2−2a+6a2−2a+6 is min when a2−2aa2−2a is minimum i.e, a=1

Hence p2+q2p2+q2= 6-1=5.

Hope it helps you..

Answer 2

Answer:

Sum of roots = (α - 2)

Sum of roots = (α - 2)Product of roots = - (α + 1)

Sum of roots = (α - 2)Product of roots = - (α + 1)p2 + q2 = (p + q)2 - 2pq

Sum of roots = (α - 2)Product of roots = - (α + 1)p2 + q2 = (p + q)2 - 2pq= (α - 2)2 + 2(α + 1)

Sum of roots = (α - 2)Product of roots = - (α + 1)p2 + q2 = (p + q)2 - 2pq= (α - 2)2 + 2(α + 1)= α2 - 2α + 6 = (α - 1)2 + 5

Sum of roots = (α - 2)Product of roots = - (α + 1)p2 + q2 = (p + q)2 - 2pq= (α - 2)2 + 2(α + 1)= α2 - 2α + 6 = (α - 1)2 + 5Since, minimum value of square can be zero, the minimum possible value of p2 + q2 is 5. ANSWER.

Step-by-step explanation:

I HOPE THAT IT'S HELPFUL FOR YOU.