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Consider the questions:

" if a,b,c are in G.P., then equations$ ax^2 +2bx+c = 0$ and $dx^2 + 2ex +f =0$ have a common roof if$ \frac{d}{a}, \frac{e}{b}, \frac{f}{c} $are in what sequence ( am, gm, hm etc) "

" if $ \alpha, \beta$ are roots of quadratic equation $x^2 - x + p $ and $\gamma, \delta$ be the roots of equation $x^2 -4x + q =0$. If $\alpha-\beta, \gamma,$ and $\delta$ are in GP, then the integral values of p and q are respectively, are "

How do I do questions like this? What is the fastest approach?

tryst with freedom
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1 Answers1

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Here's my attempt. Trust me, I solved it in around 5 min. only typing took a bit more time.

Q.1. Consider $ax^2+2bx+c=0$.

As $a,b,c$ are in GP, let $b=ar$ and $c=ar^2$

So, we have, $ax^2+2arx+ar^2=0$ i.e. $a(x+r)^2=0$ and hence $x=−r$ is the only root of $ax^2+2bx+c=0$.

This means, $x=−r$ is also the root of $dx^2+2ex+f=0$ and we have $dr^2−2er+f=0$. Dividing this by $ar^2$ we get $$\frac{d}{a}−\frac{2e}{ar}+\frac{f}{ar2}=0$$ $$\frac{d}{a}+\frac{f}{c}=\frac{2e}{b}$$

Hence, $\frac{d}{a},\;\frac{e}{b},\;\frac{f}{c}$ are in AP.

Q.2. This same question is given in my textbook also and I think you've got a typo in your question. Instead of $\alpha-\beta$ it should be $\alpha, \beta$. I've solved the question accordingly.

Let $(\alpha,\beta,\gamma,\delta)\equiv (a, ar, ar^2, ar^3)$.

$\alpha+\beta=1\implies a(1+r) =1\\ \gamma+\delta=4\implies ar^2(1+r)=4\\ \alpha\beta=p\implies ar^2=p\\ \gamma\delta=q\implies a^2r^5=q$

Now, by first two equations we get $(r, a)=(2, \frac{1}{3}), \;(-2, -1) $.

Hence, the integral value of $p, \;q$ are $-2, \;-32$ respectively.

SarGe
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