Question

If one root is the square of another of an equation $a{x}^2-2ibx+ic=0$, then show that $i=\frac{a{c}^2-6abc}{8b^3-a^{2}c}$. $i=\sqrt{-1}$

Answer 1

Consider the given equation, $a{x}^2-2ibx+ic=0$

Let the roots of this equation are $t$ and $t^2$,(given that one root is the square of another ).

Now,

$t\times t^2=\frac{ic}{a}$

$t^3=\frac{ic}{a}$ ……..(1)

$t+t^2=-\frac{-2ib}{a}$

$t+t^2=\frac{2ib}{a}$ ……….(2)

Taking cube both sides of equation (2), we get

${(t+t^2)}^3={\left(\frac{2ib}{a}\right)}^3$

$t^3+{\left(t^2\right)}^3+3t^2.t^2+3t.{\left(t^2\right)}^2=\frac{8.i^3.b^3}{a^3}=\frac{8.i^2.i.b^3}{a^3}$

$t^3+t^6+3t^4+3t^5=\frac{8.(-1).i.b^3}{a^3}(\because i^2=-1)$

$t^3+t^6+3t^4+3t^5=\frac{-8i.b^3}{a^3}$

From equation (1),

${\left({\left(\frac{ic}{a}\right)}^{\frac{1}{3}}\right)}^3+{\left({\left(\frac{ic}{a}\right)}^{\frac{1}{3}}\right)}^6+3{\left({\left(\frac{ic}{a}\right)}^{\frac{1}{3}}\right)}^4+3{\left({\left(\frac{ic}{a}\right)}^{\frac{1}{3}}\right)}^5=\frac{-8i.b^3}{a^3}$

$\frac{ic}{a}+{\left(\frac{ic}{a}\right)}^2+3{\left(\frac{ic}{a}\right)}^{\frac{4}{3}}+3{\left(\frac{ic}{a}\right)}^{\frac{5}{3}}=\frac{-8i{b}^3}{a^3}$

$\frac{ic}{a}+\frac{i^2{c}^2}{a^2}+\frac{3{\left(i^4\right)}^{\frac{1}{3}}{c}^{\frac{4}{3}}}{a^{\frac{4}{3}}}+\frac{3{\left(i^{4}i\right)}^{\frac{5}{3}}{c}^{\frac{5}{3}}}{a^{\frac{5}{3}}}=\frac{-8i{b}^3}{a^3}$

$\frac{ic}{a}-\frac{c^2}{a^2}+\frac{3c^{\frac{4}{3}}}{a^{\frac{4}{3}}}+\frac{3{\left(i\right)}^{\frac{5}{3}}{c}^{\frac{5}{3}}}{a^{\frac{5}{3}}}=\frac{-8i{b}^3}{a^3}(\because i^3=-1)$

$\frac{ic}{a}-\frac{c^2}{a^2}+\frac{3c^{\frac{4}{3}}}{a^{\frac{4}{3}}}+\frac{3{\left(i\right)}^{\frac{5}{3}}{c}^{\frac{5}{3}}}{a^{\frac{5}{3}}}-\frac{-8i{b}^3}{a^3}=0$

$i=\frac{a{c}^2-6abc}{8b^3-a^{2}c}$.

Hence, this is the answer.