Question

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

• A. $16x^2–12x+1=0$
• B. $16x^2+12x+1=0$
• C. $16x^2–12x–1=0$
• D. $16x^2+10x+1=0$

Answer 1

The correct option is A

$16x^2–12x+1=0$

Find the quadratic equation using the roots

Given roots are

${\sin }^{2}18°$ and ${\cos }^{2}36°$

so,

$$\begin{gathered} x^2–(si{n}^{2}18°+co{s}^{2}36°)x+(si{n}^{2}18°\times co{s}^{2}36°)=0 \\ {x}^2–\left[{\left[\right(\sqrt{5}–1)/4]}^2+{\left[\right(\sqrt{5}+1)/4]}^2\right]x+{{\left[\right(\sqrt{5}–1)/4]}^2\times \left[\right(\sqrt{5}+1)/4]}^2=0 \\ {x}^2–2[{(\sqrt{5}/4)}^2+{(1/4)}^2]x+{(1/4)}^2=0 \\ {x}^2–2(6/16)x+(1/16)=0 \\ 16x^2–12x+1=0 \end{gathered}$$

Hence the quadratic equation is $16x^2-12x+1=0$