Theory Of Equations Ques 2
- Let $a, b$ and $c$ be real number $s$ with $a \neq 0$ and let $\alpha, \beta$ be the roots of the equation $a x^2+b x+c=0$. Express the roots of $a^3 x^2+a b c x+c^3=0$ in terms of $\alpha, \beta$. (2001, 4M)
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Answer:
Correct Answer: 2.$(x=\alpha^2 \beta, \alpha \beta^2)$
Solution: Since, $a x^2+b x+c=0$ has roots $\alpha$ and $\beta$.
$\Rightarrow \alpha+\beta =-b / a $
and $\quad \alpha \beta =c / a$
Now, $\quad a^3 x^2+a b c x+c^3=0$ $\quad$ ……..(i)
On dividing the equation by $c^2$, we get
$\frac{a^3}{c^2} x^2+\frac{a b c x}{c^2}+\frac{c^3}{c^2}=0 $
$\Rightarrow a\left(\frac{a x}{c}\right)^2+b\left(\frac{a x}{c}\right)+c=0 $
$\Rightarrow \frac{a x}{c}=\alpha, \beta \text { are the roots } $
$\Rightarrow x=\frac{c}{a} \alpha, \frac{c}{a} \beta \text { are the roots }$
$\Rightarrow x=\alpha \beta \alpha, \alpha \beta \beta$ are the roots
$\Rightarrow x=\alpha^2 \beta, \alpha \beta^2$ are the roots
Divide the Eq. (i) by $a^3$, we get
$ \begin{array}{rlrl} & x^2+\frac{b}{a} \cdot \frac{c}{a} \cdot x+\left(\frac{c}{a}\right)^a & =0 \\ \Rightarrow & x^2-(\alpha+\beta) \cdot(\alpha \beta) x+(\alpha \beta)^3 & =0 \\ \Rightarrow & x^2-\alpha^2 \beta x-\alpha \beta^2 x+(\alpha \beta)^3 & =0 \\ \Rightarrow & x\left(x-\alpha^2 \beta\right)-\alpha \beta^2\left(x-\alpha^2 \beta\right) & =0 \\ \Rightarrow & \left(x-\alpha^2 \beta\right)\left(x-\alpha \beta^2\right) & =0 \end{array} $
$\Rightarrow x=\alpha^2 \beta, \alpha \beta^2$ which is the required answer.
Alternate Solution
Since, $\quad a^3 x^2+a b c x+c^3=0$
$ \begin{aligned} & \Rightarrow x=\frac{-a b c \pm \sqrt{(a b c)^2-4 \cdot a^3 \cdot c^3}}{2 a^3} \\ & \Rightarrow x=\frac{-(b k a)(c k a) \pm \sqrt{(b i a)^2(c k)^2-4(c a)^3}}{2} \end{aligned} $
$ \begin{aligned} & \Rightarrow x=\frac{(\alpha+\beta)(\alpha \beta) \pm \sqrt{(\alpha+\beta)^2(\alpha \beta)^2-4(\alpha \beta)^3}}{2} \\ & \Rightarrow x=\frac{(\alpha+\beta)(\alpha \beta) \pm \alpha \beta \sqrt{(\alpha+\beta)^2-4 \alpha \beta}}{2} \\ & \Rightarrow x=\alpha \beta\left[\frac{(\alpha+\beta) \pm \sqrt{(\alpha-\beta)^2}}{2}\right] \\ & \Rightarrow x=\alpha \beta\left[\frac{(\alpha+\beta) \pm(\alpha-\beta)}{2}\right] \end{aligned} $
$\Rightarrow x=\alpha \beta\left[\frac{\alpha+\beta+\alpha-\beta}{2}, \frac{\alpha+\beta-\alpha+\beta}{2}\right]$
$\Rightarrow x=\alpha \beta\left[\frac{2 \alpha}{2}, \frac{2 \beta}{2}\right]$
$\Rightarrow x=\alpha^2 \beta, \alpha \beta^2$ which is the required answer.
BETA
