Let \alpha,\ \beta be the roots of x^{2}-x+p=0 and \gamma,\

vacinammo288

vacinammo288

Answered question

2022-04-27

Let α, β be the roots of x2x+p=0 and γ, δ be the roots of x24x+q=0
where p and q are integers. IfNKSα, β, γ, δ are in GP then p+q is?

Answer & Explanation

despescarwh9

despescarwh9

Beginner2022-04-28Added 15 answers

Step 1
In the first approach: when you solve for r you would have got an equation r4=4. This gives you r2=2 and r2=2. You have only considered the first case (i.e. when r2=2). If you consider the second case r2=2, then you get r=±i2 In which case you will get p+q=34. So both approaches will yield the same answers.
Step 2
Some details:
α+β=1a(1+r2)r3=1.
Likewise
γ+δ=4ar(1+r2)=4.
Thus
r4=4
This means r2=±2 Consequently a=43r or a=4r (the second case occurs when r2=2). Now we can use,
αβ=pp=a2r4=a24.
Likewise we have
q=a2r4=4a2.
So,
p+q=174a2
Now plug in the two possible values of a2=169r2=89 or a2=16r2=8 to get both the possibilities.
Step 3
In fact, from the conditions of the problem that p and q are integers, we can see that r2=2 is not a valid solution because it gives a2=89, in which case p=a24=29 Thus r2=2 is the only valid possibility. In which case
p+q=34
is the only valid answer.

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