Theory of Equations 1 Question 4
4. If $\alpha$ and $\beta$ are the roots of the equation $x^{2}-2 x+2=0$, then the least value of $n$ for which $\frac{\alpha}{\beta}{ }^{n}=1$ is
(a) 2
(b) 5
(c) 4
(d) 3
(2019 Main, 8 April I)
Show Answer
Solution:
- Given, $\alpha$ and $\beta$ are the roots of the quadratic equation,
$$ \Rightarrow \quad \begin{array}{ll} x^{2}-2 x+2=0 \\ & (x-1)^{2}+1=0 \end{array} $$
$$ \begin{array}{lc} \Rightarrow & (x-1)^{2}=-1 \\ \Rightarrow & x-1= \pm i \\ \Rightarrow & x=(1+i) \text { or }(1-i) \end{array} $$
Clearly, if $\alpha=1+i$, then $\beta=1-i$
According to the question $\frac{\alpha}{\beta}{ }^{n}=1$
$\Rightarrow \quad \frac{1+i^{n}}{1-i}=1$
$\Rightarrow \quad \frac{(1+i)(1+i)}{(1-i)(1+i)}^{n}=1 \quad$ [by rationalization]
$\Rightarrow \quad \frac{1+i^{2}+2 i}{1-i^{2}}{ }^{n}=1 \Rightarrow \frac{2 i}{2}^{n}=1 \Rightarrow i^{n}=1$
So, minimum value of $n$ is 4 .
$\left[\because i^{4}=1\right]$
Key Idea
(i) First convert the given equation in quadratic equation.
(ii) Use, Discriminant, $D=b^{2}-4 a c<0$
Given quadratic equation is
$$ \left(1+m^{2}\right) x^{2}-2(1+3 m) x+(1+8 m)=0 $$
Now, discriminant
$$ \begin{aligned} D & =[-2(1+3 m)]^{2}-4\left(1+m^{2}\right)(1+8 m) \\ & =4\left[(1+3 m)^{2}-\left(1+m^{2}\right)(1+8 m)\right] \\ & =4\left[1+9 m^{2}+6 m-\left(1+8 m+m^{2}+8 m^{3}\right)\right] \\ & =4\left[-8 m^{3}+8 m^{2}-2 m\right] \\ & =-8 m\left(4 m^{2}-4 m+1\right)=-8 m(2 m-1)^{2} \end{aligned} $$
According to the question there is no solution of the quadratic Eq. (i), then
$$ \begin{aligned} D & <0 \\ \therefore \quad-8 m(2 m-1)^{2} & <0 \quad \Rightarrow \quad m>0 \end{aligned} $$
So, there are infinitely many values of ’ $m$ ’ for which, there is no solution of the given quadratic equation.
