The quadratic equation whose roots are ² 18^{} and ² 36^{} is :

Correct Answer is : 16x2−12x+1=016 x² - 12 x + 1 = 016x2−12x+1=0

Correct Answer is : 16x2−12x+1=016 x² - 12 x + 1 = 016x2−12x+1=0

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EAMCET Engineering 2006 Solved Paper

Section: Mathematics

Question : 106 of 160

Marks: +1, -0

\sin^2 18^{\circ}

\cos^2 36^{\circ}

Solution:

Since

\sin^2 18^{\circ}

and

\cos^2 36^{\circ}

are the roots of a quadratic equation.

herefore\; ext{ Sum of roots }\; = \sin^2 18^{\circ} + \cos^2 36^{\circ}

\;= \left(\; \frac{\sqrt{5}-1}{4}\right)^2 + \left(\; \frac{\sqrt{5}+1}{4}\right)^2

\;= \;\frac{5+1-2\sqrt{5}}{16} + \;\frac{5+1+2\sqrt{5}}{16}

\;= \;\frac{12}{16} = \;\frac{3}{4}

\;\; ext{ and product of roots }\; = \sin^2 18^{\circ} \cdot \cos^2 36^{\circ}

\;= \left(\;\frac{\sqrt{5}-1}{4}\right)^2 \left(\;\frac{\sqrt{5}+1}{4}\right)^2

\;= \left(\frac{5-1}{4 \times 4}\right)^2 = \left(\frac{1}{4}\right)^2 = \frac{1}{16}

\;

Required equation whose roots are

\sin^2 18^{\circ}

and

\cos^2 36^{\circ}

, is

\;x^2 - \;\text{ (sum of roots) }\; x + \left(\;\text{ (product of roots) }\;\right) = 0

\Rightarrow x^2 - \frac{3}{4}x + \frac{1}{16} = 0

\Rightarrow 16 x^2 - 12 x + 1 = 0

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