Theory Of Equations Ques 64
- 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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Answer:
Correct Answer: 64.$x= \alpha ^2 \beta , \alpha \beta ^2$
Solution:
Formula:
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, $a^{3} x^{2}+a b c x+c^{3}=0$ ……(i)
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 \quad x=\alpha^{2} \beta, \alpha \beta^{2} \text { are the roots } \end{array} $
Divide Eq. (i) by $a^{3}$, we get
$ \begin{array}{rlrl} & & x^{2}+\frac{b}{a} \cdot \frac{c}{a} \cdot x+(\frac{c}{a})^{3} & =0 \\ \Rightarrow & & x^{2}-(\alpha+\beta) \cdot(\alpha \beta) x+(\alpha \beta)^{2} & =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^{3}+a b c x+c^{3}=0 \\ & \Rightarrow \quad x=\frac{-b c \pm \sqrt{(b c)^{2}-4 \cdot a \cdot c^{2}}}{2 a} \\ & \Rightarrow \quad x=\frac{-\left(\frac{b}{a}\right)\left(\frac{c}{a}\right) \pm \sqrt{\left(\frac{b}{a}\right)^{2}\left(\frac{c}{a}\right)^{2}-4\left(\frac{c}{a}\right)^{3}}}{2} \\ & \Rightarrow \quad x=\frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^{2}-4\alpha \beta}}{2} \\ & \Rightarrow \quad x=\frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}}{2} \\ & \Rightarrow x=\alpha \beta \quad [\frac{(\alpha+\beta) \pm |\alpha-\beta|}{2} ] & \Rightarrow \quad x=\alpha \beta [\frac{(\alpha+\beta) \pm(\alpha-\beta)}{2}] \\ & \Rightarrow \quad x=\alpha \beta [\frac{\alpha+\beta}{2}, \frac{\alpha+\beta}{2}] \\ & \Rightarrow \quad x=\alpha \beta [\alpha, \beta] \\ & \Rightarrow \quad x=\alpha^{2} \beta \text{, } \alpha \beta^{2} \text{ which is the required answer. } \end{aligned} $
BETA
