How do i approach ahead in this question Let

afasiask7xg

afasiask7xg

Answered question

2022-03-26

How do i approach ahead in this question
Let abN, a is not equal to b and the quadratic equations (a1)x2(a2+2)x+a2+2a=0 and (b1)x2(b2+2)x+b2+2b=0 have a common root, then the value of ab5 is

Answer & Explanation

smekkleg5hhp

smekkleg5hhp

Beginner2022-03-27Added 8 answers

Let qa(x)=(a1)x2(a2+2)x+a2+2a. Any root r of qa(x) and of qb(x) is also a root of
(b-1)qa(x)-(a-1)qb(x)=(-a2b+a2+ab2+2a-b2-2b)(x-1).
Therefore, r=1 or
a2b+a2+ab2+2ab22b=0 (1)
But you can't have r=1, because
1 is a root of qa(x)qa(1)=0
3a3=0
a=1.
So, if 1 was a root of both qa(x) and qb(x), you would have a=b=1, but you are assuming that ab.
If, on the other hand, you have (1), then a=b or a=b+2b1=1+3b1. Since a, bN, this can only occur in two case: when b=2 (in which case a=4) and when b=4 (in which case a=2). In both cases, you have ab5=85.

Demetrius Kaufman

Demetrius Kaufman

Beginner2022-03-28Added 10 answers

Hint: t=ab are the two roots of the quadratic ,(t1)x2(t2+2)x+t2+2t=0 for the value of x which is the common root.

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