Question

If the sum of the roots of the quadratic equation $a{{x}^{2}}+bx+c=0$ is equal to the sum of the square of their reciprocals, then prove that $2{{a}^{2}}c={{c}^{2}}b+{{b}^{2}}a$.

Answer 1

Hint: Take the roots as P and Q. Thus find the sum of roots (P+Q) and product of roots (PQ) from the quadratic equation. Then from the question find the relation connecting P and Q. Then substitute the value of (P+Q) and (PQ) and prove the given expression.

Complete step-by-step answer:
Given to us a quadratic equation,
$a{{x}^{2}}+bx+c=0\text{ }\to (1)$
A quadratic polynomial can be also written as,
${{x}^{2}}-\left( sum\text{ }of\text{ }roots \right)x+\left( product\text{ }of\text{ }roots \right)=0$
Let us consider the two roots of the quadratic equation as P and Q.
Now let us find the sum and product of roots. From (1)
$sum\text{ }of\text{ }roots=P+Q=\dfrac{-\text{ }coefficient\text{ }of\text{ }x}{coefficient\text{ }of\text{ }{{x}^{2}}}$
$\therefore P+Q=\dfrac{-b}{a}$
$Similarly\text{ }product\text{ }of\text{ }roots=PQ=\dfrac{\text{constant term}}{coefficient\text{ }of\text{ }{{x}^{2}}}$
$\therefore PQ=\dfrac{c}{a}$
Hence we got the sum of the roots as $P+Q=\dfrac{-b}{a}$.
Thus product of the roots are $PQ=\dfrac{c}{a}$.
Now according to the question, the sum of roots of the quadratic equation is equal to the sum of the square of their reciprocals.
i.e. $P+Q={{\left( \dfrac{1}{P} \right)}^{2}}+{{\left( \dfrac{1}{Q} \right)}^{2}}$
Now let us simplify this further,

& P+Q=\dfrac{1}{{{P}^{2}}}+\dfrac{1}{{{Q}^{2}}} \\\ & P+Q=\dfrac{{{P}^{2}}+{{Q}^{2}}}{{{P}^{2}}{{Q}^{2}}} \\\ & P+Q=\dfrac{{{P}^{2}}+{{Q}^{2}}}{{{\left( PQ \right)}^{2}}} \\\ \end{aligned}$$ We can write $${{P}^{2}}+{{Q}^{2}}$$ as $${{\left( P+Q \right)}^{2}}-2PQ$$. I.e. we know $${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$$. $$\therefore {{a}^{2}}+{{b}^{2}}={{\left( a+b \right)}^{2}}-2ab,$$ Now apply this in the above expression. $$P+Q=\dfrac{{{\left( P+Q \right)}^{2}}-2PQ}{{{\left( PQ \right)}^{2}}}$$ Apply cross multiplication property, $${{\left( PQ \right)}^{2}}\left( P+Q \right)={{\left( P+Q \right)}^{2}}-2PQ$$ Now put $$P+Q=\dfrac{-b}{a},$$and $$PQ=\dfrac{c}{a}$$ $${{\left( \dfrac{c}{a} \right)}^{2}}\left( -\dfrac{b}{a} \right)={{\left( -\dfrac{b}{a} \right)}^{2}}-\left( \dfrac{2\times c}{a} \right)$$ Now simplify the above expression $$\begin{aligned} & -\dfrac{b{{c}^{2}}}{{{a}^{2}}}=\dfrac{{{b}^{2}}}{{{a}^{2}}}-\dfrac{2c}{a} \\\ & \dfrac{2c}{a}=\dfrac{{{b}^{2}}}{{{a}^{2}}}+\dfrac{b{{c}^{2}}}{{{a}^{2}}} \\\ & \dfrac{2c}{a}=\dfrac{b}{{{a}^{2}}}\left[ b+\dfrac{{{c}^{2}}}{{{a}^{2}}} \right] \\\ & 2c=\dfrac{a\times b}{{{a}^{2}}}\left[ b+\dfrac{{{c}^{2}}}{{{a}^{2}}} \right]=\dfrac{b}{a}\left[ b+\dfrac{{{c}^{2}}}{{{a}^{2}}} \right] \\\ & 2ca=b\left[ b+\dfrac{{{c}^{2}}}{{{a}^{2}}} \right],\text{ let us simplify this further}\text{.} \\\ & 2ca=\dfrac{b\left[ ab+{{c}^{2}} \right]}{a} \\\ & 2c{{a}^{2}}=a{{b}^{2}}+b{{c}^{2}} \\\ \end{aligned}$$ Hence we proved that $$2{{a}^{2}}c={{c}^{2}}b+{{b}^{2}}a$$. $$\therefore $$ The sum of the roots of the quadratic equation is equal to the sum of the square of the reciprocal and we product $$2{{a}^{2}}c={{c}^{2}}b+{{b}^{2}}a$$. Note: We know that the roots of a quadratic equation can also be found as $$P=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}\text{ }Q=\dfrac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$$ $$\therefore P+Q=\dfrac{-b+\sqrt{{{b}^{2}}-4ac}-b-\sqrt{{{b}^{2}}-4ac}}{2a}=\dfrac{-2b}{2a}=\dfrac{-b}{a}.$$ $$PQ=\dfrac{\left( -b+\sqrt{{{b}^{2}}-4ac} \right)\left( -b-\sqrt{{{b}^{2}}-4ac} \right)}{4{{a}^{2}}}=\dfrac{{{\left( -b \right)}^{2}}-{{\left( \sqrt{{{b}^{2}}-4ac} \right)}^{2}}}{4{{a}^{2}}}=\dfrac{{{b}^{2}}-{{b}^{2}}+4ac}{4{{a}^{2}}}=\dfrac{c}{a}$$.But it is difficult and lengthy process so use direct formula of finding sum and product of roots.