Question

If roots of the quadratic equation $ax^{2}+bx+c=0$ are $\alpha ,\beta$ and $3b^{2}=16ac$, then
A) $\alpha =4\beta$ or $\beta =4\alpha$
B) $\alpha =-4\beta$ or $\beta =-4\alpha$
C) $\alpha =3\beta$ or $\beta =3\alpha$
D) $\alpha =-3\beta$ or $\beta =-3\alpha$

Answer 1

Hint: In this question it is given that $\alpha ,\beta$ are the roots of the quadratic equation $ax^{2}+bx+c=0$ and $3b^{2}=16ac$, then we have to find the relation between $\alpha$ and $\beta$. So to find the solution we need to know that if $ax^{2}+bx+c=0$ be any quadratic equation and if $\alpha ,\beta$ be two roots of that equation, the we can write $$\alpha +\beta =\dfrac{-b}{a}$$ and $$\alpha \beta =\dfrac{c}{a}$$.
Complete step-by-step solution:
So by above formula we can write the relation between the roots $\alpha ,\beta$ for the quadratic equation $ax^{2}+bx+c=0$ is,
$$\alpha +\beta =\dfrac{-b}{a}$$.....(1)
$$\alpha \beta =\dfrac{c}{a}$$.........(2)
Now, dividing both side of the given equation $$3b^{2}=16ac$$ by $a^{2}$, we get,
$$3b^{2}=16ac$$
$$\Rightarrow \dfrac{3b^{2}}{a^{2}} =16\dfrac{ac}{a^{2}}$$
$$\Rightarrow 3\left( \dfrac{b}{a} \right)^{2} =16\dfrac{c}{a}$$
$$\Rightarrow 3\left( \dfrac{-b}{a} \right)^{2} =16\dfrac{c}{a}$$
$$\Rightarrow 3\left( \alpha +\beta \right)^{2} =16\alpha \beta$$ [by using (1) and (2)]
Now, as we know that $$\left( a+b\right)^{2} =a^{2}+2ab+b^{2}$$, so by this identity the above equation can be written as,
$$3\left( \alpha^{2} +2\alpha \beta +\beta^{2} \right) =16\alpha \beta$$
$$\Rightarrow 3\left( \dfrac{\alpha^{2} }{\alpha \beta } +\dfrac{2\alpha \beta }{\alpha \beta } +\dfrac{\beta^{2} }{\alpha \beta } \right) =\dfrac{16\alpha \beta }{\alpha \beta }$$ [ dividing both side by $\alpha \beta$]
$$\Rightarrow 3\left( \dfrac{\alpha }{\beta } +2+\dfrac{\beta }{\alpha } \right) =16$$
$$\Rightarrow 3\dfrac{\alpha }{\beta } +6+3\dfrac{\beta }{\alpha } =16$$
$$\Rightarrow 3\dfrac{\alpha }{\beta } +3\dfrac{\beta }{\alpha } =16-6$$
$$\Rightarrow 3\left( \dfrac{\alpha }{\beta } \right) +3\dfrac{1}{\left( \dfrac{\alpha }{\beta } \right) } =10$$........ (3)
Now let $$y=\dfrac{\alpha }{\beta }$$,
Therefore, equation (3) can be written as,
$$3y+\dfrac{3}{y} =10$$
$$\Rightarrow 3y^{2}+3=10y$$
$$\Rightarrow 3y^{2}-10y+3=0$$.......(4)
Now as we know that if any equation is in the form of $ay^{2}+by+c=0$, then by quadratic formula ‘y’ can be written as,
$$y=\dfrac{-b\pm \sqrt{b^{2}-4ac} }{2a}$$
Now if we compare equation (4) with $ay^{2}+by+y=0$, then we can write, a=3, b=-10, c=3, and
$$y=\dfrac{-\left( -10\right) \pm \sqrt{\left( -10\right)^{2} -4\cdot 3\cdot 3} }{2\cdot 3}$$
$$\Rightarrow y=\dfrac{10\pm \sqrt{100-36} }{6}$$
$$\Rightarrow y=\dfrac{10\pm \sqrt{64} }{6}$$
$$\Rightarrow y=\dfrac{10\pm 8}{6}$$
$$\therefore y=\dfrac{10+8}{6} ,\ \dfrac{10-8}{6}$$
$$\Rightarrow y=\dfrac{18}{6} ,\ \dfrac{2}{6}$$
$$\Rightarrow y=3,\ \dfrac{1}{3}$$
Therefore, by putting the value of ‘y’, we get,
  $$\dfrac{\alpha }{\beta } =3\ \text{or} \ \dfrac{\alpha }{\beta } =\dfrac{1}{3}$$
$$\Rightarrow \alpha =3\beta \ \text{or} \ \text{and} \ \beta =3\alpha$$
Hence, the correct option is option C.
Note: While solving this question we used the formula where the relation between the two roots $\alpha$ and $\beta$ are there, because option is given like the relation between the $\alpha$ and $\beta$, so that is why we use that relation formula.
but you can also use the quadratic formula but this will lead to a clumsy and lengthy solution.