Question

Show that the equation px² + qx + r = 0 and qx² + rx + p = 0 will have a common root if p+q+r=0 or p=q=r.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given pair of quadratic equations are

$\rm :\longmapsto\:p {x}^{2} + qx + r = 0 - - - (1)$

and

$\rm :\longmapsto\:q{x}^{2} + rx + p= 0 - - - (2)$

Let assume that the common root be 'y'.

So, above equations can be rewritten as

$\rm :\longmapsto\:p {y}^{2} + qy + r = 0 - - - (3)$

and

$\rm :\longmapsto\:q{y}^{2} + ry + p= 0 - - - (4)$

So, Using Cross Multiplication method we have

$\begin{gathered}\boxed{\begin{array}{c|c|c|c} \bf 2 & \bf 3 & \bf 1& \bf 2\\ \frac{\qquad}{} & \frac{\qquad}{}\frac{\qquad}{} &\frac{\qquad}{} & \frac{\qquad}{} &\\ \sf q & \sf r & \sf p & \sf q\\ \\ \sf r & \sf p & \sf q & \sf r\\ \end{array}} \\ \end{gathered}$

So,

$\rm :\longmapsto\:\dfrac{ {y}^{2} }{pq - {r}^{2} } = \dfrac{y}{ {p}^{2} - rq} = \dfrac{1}{pr - {q}^{2} }$

$\rm :\implies\:y = \dfrac{ {p}^{2} - rq }{pr - {q}^{2} }$

and

$\rm :\implies\: {y}^{2} = \dfrac{pq - {r}^{2} }{pr - {q}^{2} }$

On Substituting the value of y, we get

$\rm :\implies\: \dfrac{ {( {p}^{2} - rq) }^{2} }{ {(pr - {q}^{2} )}^{2} } = \dfrac{pq - {r}^{2} }{pr - {q}^{2} }$

$\rm :\implies\: \dfrac{ {( {p}^{2} - rq) }^{2} }{ {(pr - {q}^{2} )}} = pq - {r}^{2}$

$\rm :\implies\: {( {p}^{2} - rq) }^{2} = (pq - {r}^{2}) (pr - {q}^{2} )$

$\rm :\longmapsto\: {p}^{4} + {r}^{2} {q}^{2} - 2 {p}^{2}qr = {p}^{2}qr - {pr}^{3} - {pq}^{3} + {q}^{2} {r}^{2}$

$\rm :\longmapsto\: {p}^{4} - 3{p}^{2}qr + {pr}^{3} + {pq}^{3} = 0$

$\rm :\longmapsto\: p({p}^{3}+{r}^{3} + {q}^{3} - 3pqr) = 0$

$\rm :\longmapsto\: {p}^{3}+{r}^{3} + {q}^{3} - 3pqr = 0$

$\rm :\longmapsto\:(p + q + r)( {p}^{2} + {q}^{2} + {r}^{2} - pq - qr - rp) = 0$

$\bf\implies \:p + q + r = 0$

or

$\bf\implies \:{p}^{2} + {q}^{2} + {r}^{2} - pq - qr - rp = 0$

$\rm :\longmapsto\: \:2{p}^{2} + 2{q}^{2} + 2{r}^{2} -2 pq -2 qr - 2rp= 0$

$\rm :\longmapsto\:{p}^{2} + {p}^{2}+{q}^{2}+{q}^{2}+{r}^{2} + {r}^{2} -2 pq -2 qr - 2rp= 0$

$\rm :\longmapsto\:( {p}^{2} + {q}^{2} - 2pq) + ( {q}^{2} + {r}^{2} - 2qr) + ( {r}^{2} + {p}^{2} - 2pr) = 0$

$\rm :\longmapsto\: {(p - q)}^{2} + {(q - r)}^{2} + {(r - p)}^{2} = 0$

We know, Sum of squares is 0 only, when

$\rm :\longmapsto\: {(p - q)} = 0 \: and \: {(q - r)} = 0 \: and \:{(r - p)} = 0$

$\rm :\longmapsto\:p = q \: and \: q = r \: and \: r = p$

$\bf\implies \:p = q = r$

Hence, Proved