The quadratic equation whose roots are sin ^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4

Vedclass · Accepted Answer

$16 x^2-12 x+1=0$

Vedclass · Answer

(C) We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.$\sin ^2 18^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 = \frac{5+1-2\sqrt{5}}{16} = \frac{6-2\sqrt{5}}{16} = \frac{3-\sqrt{5}}{8}$.$\cos ^2 36^{\circ} = \left(\frac{\sqrt{5}+1}{4}ight)^2 = \frac{5+1+2\sqrt{5}}{16} = \frac{6+2\sqrt{5}}{16} = \frac{3+\sqrt{5}}{8}$.The sum of the roots is $\frac{3-\sqrt{5}}{8} + \frac{3+\sqrt{5}}{8} = \frac{6}{8} = \frac{3}{4}$.The product of the roots is $\left(\frac{3-\sqrt{5}}{8}ight) \left(\frac{3+\sqrt{5}}{8}ight) = \frac{9-5}{64} = \frac{4}{64} = \frac{1}{16}$.The quadratic equation is given by $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - \frac{3}{4}x + \frac{1}{16} = 0$.Multiplying by $16$,we get $16x^2 - 12x + 1 = 0$.

The quadratic equation whose roots are $\sin ^2 18^{\circ}$ and $\cos ^2 36^{\circ}$ is

Similar Questions

The solution set of the congruence $8x \equiv 6 \pmod{14}$,where $x \in \mathbb{Z}$,is:

If a polynomial $f(x)$ is divided by $(x + 1)$,$(x - 2)$,and $(x + 2)$,the remainders are $6$,$3$,and $15$ respectively. Find the remainder when $f(x)$ is divided by $(x + 1)(x - 2)(x + 2)$.

Let $S = \{ x : x \in R \text{ and } (\sqrt{3} + \sqrt{2})^{x^2 - 4} + (\sqrt{3} - \sqrt{2})^{x^2 - 4} = 10 \}$. Then $n(S)$ is equal to

The number of rational roots of the equation $x^{2016} - x^{2015} + x^{1008} + x^{1003} + 1 = 0$ is equal to

The roots of the given equation $2(a^2 + b^2)x^2 + 2(a + b)x + 1 = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module