Let&nbsp;α,&nbsp;β&nbsp;be the roots of&nbsp;x2−x+p=0&nbsp;and&nbsp;γ,&nbsp;δ&nbsp;be the roots of&nbsp;x2−4x+q=0where p and q are integers. If&nbsp;α,&nbsp;β,&nbsp;γ,&nbsp;δ&nbsp;are in GP then&nbsp;p+q&nbsp;is?

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Let&amp;nbsp;α,&amp;nbsp;β&amp;nbsp;be the roots of&amp;nbsp;x2−x+p=0&amp;nbsp;and&amp;nbsp;γ,&amp;nbsp;δ&amp;nbsp;be the roots of&amp;nbsp;x2−4x+q=0where p and q are integers. If&amp;nbsp;α,&amp;nbsp;β,&amp;nbsp;γ,&amp;nbsp;δ&amp;nbsp;are in GP then&amp;nbsp;p+q&amp;nbsp;is?

Ireland Vaughan · Accepted Answer

Step 1In the first approach: when you solve for r you would have got an equation&amp;nbsp;r4=4. This gives you&amp;nbsp;r2=2&amp;nbsp;and&amp;nbsp;r2=−2. You have only considered the first case (i.e. when&amp;nbsp;r2=2). If you consider the second case&amp;nbsp;r2=−2, then you get&amp;nbsp;r=±i2&amp;nbsp;In which case you will get&amp;nbsp;p+q=−34. So both approaches will yield the same answers.Step 2Some details:α+β=1⇒a(1+r2)r3=1.Likewiseγ+δ=4⇒ar(1+r2)=4.Thusr4=4This means&amp;nbsp;r2=±2&amp;nbsp;Consequently&amp;nbsp;a=43r&amp;nbsp;or&amp;nbsp;a=−4r&amp;nbsp;(the second case occurs when&amp;nbsp;r2=−2). Now we can use,α⋅β=p⇒p=a2r4=a24.Likewise we haveq=a2r4=4a2.So,p+q=174a2Now plug in the two possible values of&amp;nbsp;a2=169r2=89&amp;nbsp;or&amp;nbsp;a2=16r2=−8&amp;nbsp;to get both the possibilities.Step 3In fact, from the conditions of the problem that p and q are integers, we can see that&amp;nbsp;r2=2&amp;nbsp;is not a valid solution because it gives&amp;nbsp;a2=89, in which case&amp;nbsp;p=a24=29∈ℤ&amp;nbsp; Thus&amp;nbsp;r2=−2&amp;nbsp;is the only valid possibility. In which casep+q=−34is the only valid answer.

Let \(\displaystyle\alpha,\ \beta\) be the roots of

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