Question

If $\alpha ,\beta$ are real and $$\alpha^{2} ,-\beta^{2}$$ are the roots of the quadratic equation $$a^{2}x^{2}+x+1-a^{2}=0$$;(a > 1), then $\beta^{2} =$
A) $$a^{2}$$
B) 1
C) $$1-a^{2}$$
D) $$1+a^{2}$$

Answer 1

Hint: In this question it is given that $$\alpha^{2} ,-\beta^{2}$$ are the roots of $$a^{2}x^{2}+x+1-a^{2}=0$$;(a > 1), then we have to find the value of $$\beta^{2}$$. So for this we have to know that if p and q are the root of the quadratic equation $rx^{2}+sx+t=0$, then we can write, $$p+q=\dfrac{-s}{r}$$ and $$pq=\dfrac{t}{r}$$

Complete step-by-step solution:
Given equation $$a^{2}x^{2}+x+1-a^{2}=0$$.......(1)
So comparing equation(1) with $rx^{2}+sx+t=0$ we can write,
r=$a^{2}$, s=1, t=$1-a^{2}$
And the roots are $$\alpha^{2} ,-\beta^{2}$$.
Therefore, by the above formula we can write,
$$\alpha^{2} +\left( -\beta^{2} \right) =\dfrac{-1}{a^{2}}$$
$$\Rightarrow \alpha^{2} -\beta^{2} =-\dfrac{1}{a^{2}}$$........(2)
And, $$\alpha^{2} \left( -\beta^{2} \right) =\dfrac{1-a^{2}}{a^{2}}$$
  $$\Rightarrow \alpha^{2} \beta^{2} =\dfrac{a^{2}-1}{a^{2}}$$
  $$\Rightarrow \alpha^{2} =\dfrac{a^{2}-1}{\beta^{2} a^{2}}$$.......(3)
Now putting the value of $\alpha^{2}$ in the equation (2), we get,
$$\dfrac{a^{2}-1}{\beta^{2} a^{2}} -\beta^{2} =-\dfrac{1}{a^{2}}$$
Let $$\beta^{2}=y$$, therefore, the above equation can be written as,
$$\dfrac{a^{2}-1}{ya^{2}} -y=-\dfrac{1}{a^{2}}$$
$$\Rightarrow a^{2}-1-y^{2}a^{2}=-\dfrac{1}{a^{2}} \times a^{2}y$$[multiplying both side by $$ya^{2}$$]
$$\Rightarrow a^{4}-a^{2}-y^{2}a^{4}=-a^{2}y$$[multiplying both side by $$a^{2}$$]
$$\Rightarrow a^{4}y^{2}-a^{2}y+a^{2}-a^{4}=0$$
$$\Rightarrow a^{4}y^{2}-a^{2}y+(a^{2}-a^{4})=0$$......(4)
Therefore, by quadratic formula,
$$y=\dfrac{-\left( -a^{2}\right) \pm \sqrt{\left( -a^{2}\right)^{2} -4\cdot a^{4}\cdot \left( a^{2}-a^{4}\right) } }{2a^{4}}$$
$$=\dfrac{a^{2}\pm \sqrt{a^{4}-4a^{6}+4a^{8}} }{2a^{4}}$$
$$=\dfrac{a^{2}\pm \sqrt{\left( a^{2}\right)^{2} -2\cdot a^{2}\cdot \left( 2a^{4}\right) +\left( 2a^{4}\right)^{2} } }{2a^{4}}$$
Since as we know that, $$x^{2}-2xy+y^{2}=\left( x-y\right)^{2} $$, so by using the identity we can write the above equation as,
$$y=\dfrac{a^{2}\pm \sqrt{\left( a^{2}-2a^{4}\right)^{2} } }{2a^{4}}$$
$$y=\dfrac{a^{2}\pm \left( a^{2}-2a^{4}\right) }{2a^{4}}$$
$$\ \text{Either,} \ y=\dfrac{a^{2}+\left( a^{2}-2a^{4}\right) }{2a^{4}} \ \text{or} \ y=\dfrac{a^{2}-\left( a^{2}-2a^{4}\right) }{2a^{4}}$$
Therefore,
$$y=\dfrac{2a^{2}-2a^{4}}{2a^{4}} \ \text{or} \ y=\dfrac{2a^{4}}{2a^{4}}$$
$$\Rightarrow y=\dfrac{2a^{2}\left( 1-a^{2}\right) }{2a^{4}} \ \text{or} \ y=1$$
$$\Rightarrow y=\dfrac{\left( 1-a^{2}\right) }{a^{2}} \ \text{or} \ y=1$$
$$\Rightarrow \beta^{2} =\dfrac{\left( 1-a^{2}\right) }{a^{2}} \ \text{or} \ \beta^{2} =1$$ [ since, $$y=\beta^{2}$$]
Therefore, the second value of $\beta^{2}$ is correct, i,e $$\beta^{2} =1$$.
Hence the correct option is option B.

Note: So you can also solve this in different way, since $-\beta^{2}$ is the root of the given equation $a^{2}x^{2}+x+1-a^{2}=0$, so you $-\beta^{2}$ must be satisfies the equation, i.e, you can put $-\beta^{2}$ in the place of ‘x’, which gives a quadratic equation of $\beta^{2}$ and after solving can able to find the solution.