Question

If sin a and cos a are the roots of the equation 4x^2 -kx-1 =0 (k>0) then the value of k is

Answer 1

Answer:

$\sf{The \ value \ of \ k \ is \ 2\sqrt2.}$

Given:

• The given quadratic equation is
• $\sf{4x^{2}-kx-1=0}$

• Roots are sin a and cos a

To find:

• The value of k.

Solution:

$\sf{The \ given \ quadratic \ equation \ is}$

$\sf{4x^{2}-kx-1=0}$

$\sf{Here, \ a=4, \ b=-k \ and \ -1}$

$\sf{Sum \ of \ roots=\dfrac{-b}{a}}$

$\sf{\therefore{sin \ a+cos \ a=\dfrac{k}{4}...(1)}}$

$\sf{Product \ of \ roots=\dfrac{c}{a}}$

$\sf{\therefore{sin \ a\times \ cos \ a=\dfrac{-1}{4}...(2)}}$

$\sf{According \ to \ the \ Trigonometric \ identity}$

$\sf{sin^{2}\theta+cos^{2}\theta=1}$

$\sf{\therefore{sin^{2}a+cos^{2}a=1...(3)}}$

$\sf{According \ to \ the \ identity.}$

$\boxed{\sf{a^{2}+b^{2}=(a+b)^{2}-2ab}}$

$\sf{\therefore{sin \ a^{2}+cos^{2} \ a=(sin \ a+cos \ a)^{2}-2(sin \ a.cos \ a)}}$

$\sf{...from \ (1), \ (2) \ and \ (3)}$

$\sf{1=(\dfrac{k}{4})^{2}-2(\dfrac{-1}{4})}$

$\sf{\therefore{1=\dfrac{k^{2}}{16}+\dfrac{1}{2}}}$

$\sf{Multiply \ throughout \ by \ 16 \ we \ get,}$

$\sf{16=k^{2}+8}$

$\sf{\therefore{k^{2}=16-8}}$

$\sf{\therefore{k^{2}=8}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\sf{\therefore{k=\pm \ 2\sqrt2}}$

$\sf{But, \ k \ is \ bigger \ than \ zero.}$

$\sf{\therefore{k \ \neq \ -2\sqrt2}}$

$\sf{\therefore{k=2\sqrt2}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ k \ is \ 2\sqrt2.}}}$