Theory of Equations 3 Question 2
2. A value of $b$ for which the equations $x^{2}+b x-1=0$, $x^{2}+x+b=0$ have one root in common is
(2011)
(a) $-\sqrt{2}$
(b) $-i \sqrt{3}$
(c) $i \sqrt{5}$
(d) $\sqrt{2}$
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Solution:
- Since, $a x^{2}+b x+c=0$ has roots $\alpha$ and $\beta$.
$$ \begin{aligned} \Rightarrow & \alpha+\beta & =-b / a \\ \text { and } & \alpha \beta & =c / a \end{aligned} $$
Now, $a^{3} x^{2}+a b c x+c^{3}=0$
On dividing the equation by $c^{2}$, we get
$$ \begin{array}{ll} & \frac{a^{3}}{c^{2}} x^{2}+\frac{a b c x}{c^{2}}+\frac{c^{3}}{c^{2}}=0 \\ \Rightarrow \quad & a \frac{a x}{c}+b \frac{a x}{c}+c=0 \\ \Rightarrow \quad & \frac{a x}{c}=\alpha, \beta \text { are the roots } \\ \Rightarrow \quad & x=\frac{c}{a} \alpha, \frac{c}{a} \beta \text { are the roots } \end{array} $$
$$ \begin{array}{ll} \Rightarrow & x=\alpha \beta \alpha, \alpha \beta \beta \text { are the roots } \\ \Rightarrow & x=\alpha^{2} \beta, \alpha \beta^{2} \text { are the roots } \end{array} $$
Divide the Eq. (i) by $a^{3}$, we get
$$ \begin{array}{rlrl} & & x^{2}+\frac{b}{a} \cdot \frac{c}{a} \cdot x+\frac{c}{a} & =0 \\ \Rightarrow & & x^{2}-(\alpha+\beta) \cdot(\alpha \beta) x+(\alpha \beta)^{3} & =0 \\ \Rightarrow & x^{2}-\alpha^{2} \beta x-\alpha \beta^{2} x+(\alpha \beta)^{3} & =0 \\ \Rightarrow & x\left(x-\alpha^{2} \beta\right)-\alpha \beta^{2}\left(x-\alpha^{2} \beta\right) & =0 \\ \Rightarrow & & \left(x-\alpha^{2} \beta\right)\left(x-\alpha \beta^{2}\right) & =0 \end{array} $$
$\Rightarrow x=\alpha^{2} \beta, \alpha \beta^{2}$ which is the required answer.
Alternate Solution
$$ \begin{aligned} & \text { Since, } \quad a^{3} x^{2}+a b c x+c^{3}=0 \\ & \Rightarrow \quad x=\frac{-a b c \pm \sqrt{(a b c)^{2}-4 \cdot a^{3} \cdot c^{3}}}{2 a^{3}} \\ & \Rightarrow \quad x=\frac{-(b / a)(c / a) \pm \sqrt{(b / a)^{2}(c / a)^{2}-4(c / a)^{3}}}{2} \\ & \Rightarrow \quad x=\frac{(\alpha+\beta)(\alpha \beta) \pm \sqrt{(\alpha+\beta)^{2}(\alpha \beta)^{2}-4(\alpha \beta)^{3}}}{2} \\ & \Rightarrow \quad x=\frac{(\alpha+\beta)(\alpha \beta) \pm \alpha \beta \sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}}{2} \\ & \Rightarrow x=\alpha \beta \quad \frac{(\alpha+\beta) \pm \sqrt{(\alpha-\beta)^{2}}}{2} \\ & \Rightarrow \quad x=\alpha \beta \frac{(\alpha+\beta) \pm(\alpha-\beta)}{2} \\ & \Rightarrow \quad x=\alpha \beta \frac{\alpha+\beta+\alpha-\beta}{2}, \frac{\alpha+\beta-\alpha+\beta}{2} \\ & \Rightarrow \quad x=\alpha \beta \frac{2 \alpha}{2}, \frac{2 \beta}{2} \\ & \Rightarrow \quad x=\alpha^{2} \beta, \alpha \beta^{2} \text { which is the required answer. } \end{aligned} $$
