Question

36 If p, q and r are in GP and the equation px2 + 2qx +r =0 and dx2 +2ex + f =0 have a common root, then show that d/p, e/q, f/r are in A.P.

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Answer 1

Step-by-step explanation :

Given,

• p , q and r are in G.P.

we know,

In G.P., the ratio between any two consecutive terms is constant.

Therefore,

    $\bf \frac{q}{p} =\frac{r}{q} \\\\ q \times q=r \times p \\\\ \boxed{\bf q^2=rp}$

____________________________

Given equation,

 px² + 2qx + r = 0

It is of the form ax² + bx + c = 0

a = p , b = 2q , c = r

we know,

        $\boxed{\bf x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

Substitute the values of a, b and c

        $\bf x=\frac{-2q\pm\sqrt{(2q)^2-4(p)(r)}}{2(p)} \\\\\\ x=\frac{-2q\pm\sqrt{4q^2-4pr}}{2p} \\\\\\ x=\frac{-2q\pm\sqrt{4(q^2-pr)} }{2p} \\\\\\ x=\frac{-2q\pm2\sqrt{q^2-pr}}{2p} \\\\\\ x=\frac{-q\pm\sqrt{q^2-pr}}{p} \\\\\\ x=\frac{-q\pm\sqrt{pr-pr}}{p} \\\\\\ x=\frac{-q\pm\sqrt{0}}{p} \\\\\\ x=\frac{-q\pm0}{p} \\\\\\ \boxed{\bf x=\frac{-q}{p} }$

__________________________

Also given,

  px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root.

So, -q/p is also a root of dx² + 2ex + f = 0

Substitute x = -q/p in the equation dx² + 2ex + f = 0

            dx² + 2ex + f = 0

           $\bf d(\frac{-q}{p})^2+2e(\frac{-q}{p})+f=0 \\\\\\ d(\frac{q^2}{p^2})-\frac{2eq}{p}+f=0 \\\\\\ \frac{dq^2}{p^2}-\frac{2eq}{p}+f=0 \\\\\\ \frac{dq^2-2eqp+fp^2}{p^2} =0 \\\\\\ dq^2-2eqp+fp^2=0$

Divide this equation with pq²,

            $\bf \frac{dq^2}{pq^2} -\frac{2eqp}{pq^2}+\frac{fp^2}{pq^2}=0 \\\\\\\ \frac{d}{p} -\frac{2e}{q} +\frac{fp}{q^2}=0 \\\\\\ \frac{d}{p} +\frac{fp}{q^2}=\frac{2e}{q} \\\\\\ \rightarrow Put \ q^2=rp, \\\\ \frac{d}{p} +\frac{fp}{rp}=\frac{2e}{q} \\\\\\ \frac{d}{p}+\frac{f}{r}=\frac{2e}{q} \\\\\\ \frac{d}{p}+\frac{f}{r} =\frac{e}{q}+\frac{e}{q} \\\\\\ \frac{e}{q}-\frac{d}{p}=\frac{f}{r}-\frac{e}{q}$

Since, the difference between the consecutive terms is constant.

    $\bf \frac{d}{p},\frac{e}{q} \ and \ \frac{f}{r} \ are \ in \ A.P.$

Hence proved !!