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

(A) Given roots are $\alpha = \sin ^2 18^{\circ}$ and $\beta = \cos ^2 36^{\circ}$.We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Given roots are $\alpha = \sin ^2 18^{\circ}$ and $\beta = \cos ^2 36^{\circ}$.We know that $\sin 18^{\circ} = \frac{\sqrt{5}-1}{4}$ and $\cos 36^{\circ} = \frac{\sqrt{5}+1}{4}$.Sum of roots $= \sin ^2 18^{\circ} + \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 + \left(\frac{\sqrt{5}+1}{4}ight)^2$.$= \frac{5+1-2\sqrt{5}}{16} + \frac{5+1+2\sqrt{5}}{16} = \frac{12}{16} = \frac{3}{4}$.Product of roots $= \sin ^2 18^{\circ} \cdot \cos ^2 36^{\circ} = \left(\frac{\sqrt{5}-1}{4}ight)^2 \cdot \left(\frac{\sqrt{5}+1}{4}ight)^2$.$= \left(\frac{(\sqrt{5}-1)(\sqrt{5}+1)}{16}ight)^2 = \left(\frac{5-1}{16}ight)^2 = \left(\frac{4}{16}ight)^2 = \left(\frac{1}{4}ight)^2 = \frac{1}{16}$.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

$A$ network of four capacitors with capacitances $C_1 = C$,$C_2 = 2C$,$C_3 = 3C$,and $C_4 = 4C$ is connected to a battery of potential $V$ as shown in the figure. The ratio of the charges on $C_2$ and $C_4$ is

$NH_3 + O_2 \xrightarrow[\Delta]{Pt} A + H_2O;$
$A + O_2 \to B$
$B + O_2 + H_2O \to C$
$A, B$ and $C$ are:

Which of the following is a linear molecule?

The point $P$ is equidistant from $A(1, 3)$,$B(-3, 5)$,and $C(5, -1)$. Then $PA =$

In which of the following reactions is the product an ether?

Vedclass Test Series

Exam Paper Generator

Online Exam Module