Question

Consider the quadratic equation $4{\cos }^{2}x+2\cos x-1=0.$
Then which of the following quadratic equation has same roots as that of the given equation is

• A. $4{\cos }^{2}2x+2\cos 2x-1=0$
• B. ${\cos }^{2}2x+2\cos 2x-4=0$
• C. $4{\cos }^{2}2x-2\cos 2x+1=0$
• D. $4{\cos }^{2}2x+\cos 2x-1=0$

Answer 1

The correct option is A $4{\cos }^{2}2x+2\cos 2x-1=0$

$4{\cos }^{2}x+2\cos x-1=0\to$
This is a quadratic equation in $\cos x$ and in all options given are quadratic equations are in $\cos 2x.$
Let the roots of the given equation be $m$ and $n$.
Then the roots of the required equation be $2m^2-1$ and $2n^2-1$ as $\cos 2x=2{\cos }^{2}x-1$.

Now,
$4{\cos }^{2}x+2\cos x-1=0$
$m+n=-\frac{1}{2},mn=-\frac{1}{4}$
Sum and product of the roots of required equation are
$2(m^2+n^2)-2=2\left[\right(m+n{)}^2-2mn]-2=-\frac{1}{2}$
Similarly,
$(2m^2-1)(2n^2-1)=4m^2{n}^2-2(m^2+n^2)+1=-\frac{1}{4}$

Therefore, required equations are
$4{\cos }^{2}2x+2\cos 2x-1=0$