Question

A quadratic equation whose roots are $si{n}^2{18}^{\circ },co{s}^2{36}^{\circ }$ are

• A. $16x^2–12x+1=0$
• B. $x^2–12x+1=0$
• C. $16x^2–2x–1=0$
• D. None

Answer 1

The correct option is A $16x^2–12x+1=0$

$Sumoftheroots=si{n}^2{18}^{\circ }+co{s}^2{36}^{\circ }$ $={\left[\frac{\sqrt{5}-1}{4}\right]}^2+{\left[\frac{\sqrt{5}-1}{4}\right]}^2=\frac{1}{16}\left[\right(\sqrt{5}+1{)}^2+\sqrt{5}-1{)}^2]=\left(\frac{1}{16}\right)\left[2\right(5+1\left)\right]=\frac{3}{4}$ $Productsoftheroots=si{n}^2{18}^{\circ }co{s}^2{36}^{\circ }={\left[\frac{\sqrt{5}-1}{4}\right]}^2{\left[\frac{\sqrt{5}+1}{4}\right]}^2={\left[\frac{5-1}{4.4}\right]}^2=\frac{1}{16}$ $\therefore QuadraticEquationis$ $x^2-[si{n}^2{18}^{\circ }+co{s}^2{36}^{\circ }]x+\left(si{n}^2{18}^{\circ }co{s}^2{36}^{\circ }\right)=0\Rightarrow x^2-\left(\frac{3}{4}\right)x+\left(\frac{1}{16}\right)=0$