Question

. If sine, cose are the roots of 6x2-px+1=0,
then p2 =​

Answer 1

$\red{\underline{\underline{Answer:}}}$

$\sf{The \ value \ of \ p^{2} \ is \ 48.}$

$\sf\orange{Given:}$

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

$\sf{\implies{6x^{2}-px+1=0}}$

$\sf{\implies{sin\theta \ and \ cos\theta \ are \ the roots}}$

$\sf\pink{To \ find:}$

$\sf{The \ value \ of \ p^{2}}$

$\sf\green{\underline{\underline{Solution:}}}$

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

$\sf{\implies{6x^{2}-px+1=0}}$

$\sf{Here \ a=6, \ b=-p \ and \ c=1}$

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

$\sf{\therefore{sin\theta+cos\theta=\frac{p}{6}...(1)}}$

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

$\sf{\therefore{sin\theta.cos\theta=\frac{1}{6}...(2)}}$

$\sf{We, \ know \ sin^{2}\theta+cos^{2}\theta=1...(3)}$

$\sf{According to the identity}$

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

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

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

$\sf{1=(\frac{p}{6})^{2}-2\times\frac{1}{6}}$

$\sf{1=\frac{p^{2}}{36}-\frac{12}{36}}$

$\sf{\therefore{p^{2}-12=36}}$

$\sf{\therefore{p^{2}=36+12}}$

$\sf{\therefore{p^{2}=48}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ p^{2} \ is \ 48.}}}$

Answer 2

$\purple{\underline{\underline{\pink{\boxed{\boxed{\red{\star{\sf Question:}}}}}}}}$

$\rm{If \ sin\theta \ and \ cos\theta \ are \ the \ roots \ of}$

$\rm{6x^{2}-px+1=0 \ then \ Find \ p^{2}}$

_________________________________________

$\purple{\underline{\underline{\orange{\boxed{\boxed{\green{\star{\sf Answer:}}}}}}}}$

$\rm{\blue{\boxed{\pink{\star{ p^{2} \ \implies \ 48.}}}}}$

_________________________________________

$\sf\large\underline\pink{Given:}$

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

$\rm{\longrightarrow{6x^{2}-px+1=0}}$

$\rm{\longrightarrow{sin\theta \ and \ cos\theta \ are \ the roots}}$

_________________________________________

$\sf\large\underline\blue{To \ Find}$

$\rm{The \ value \ of \ p^{2}}$

_________________________________________

$\green{\underline{\underline{\red{\boxed{\boxed{\purple{\star{\sf Solution:}}}}}}}}$

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

$\rm{\longrightarrow{6x^{2}-px+1=0}}$

$\rm{Here \ a=6, \ b=-p \ and \ c=1}$

$\rm{We \ know,}$

$\rm{Sum \ of \ roots=\frac{-b}{a}}$

$\rm{\implies{sin\theta+cos\theta=\frac{p}{6}...(a)}}$

$\rm{And,}$

$\rm{Product \ of \ roots=\frac{c}{a}}$

$\rm{\implies{sin\theta.cos\theta=\frac{1}{6}...(b)}}$

$\sf{We, \ know \ that}$

$\rm{ sin^{2}\theta+cos^{2}\theta=1...(c)}$

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

$\rm{sin^{2}\theta=(sin\theta+cos\theta)-2(sin\theta.cos\theta)}$

$\rm{From \ (a), \ (b) \ and (c)}$

$\rm\therefore{1=(\frac{p}{6})^{2}-2\times\frac{1}{6}}$

$\rm\therefore{1=\frac{p^{2}}{36}-\frac{12}{36}}$

$\rm{\implies{p^{2}-12=36}}$

$\rm{\implies{p^{2}=36+12}}$

$\rm{\implies{p^{2}=48}}$

$\rm{\blue{\boxed{\pink{\star{ p^{2} \ \implies \ 48.}}}}}$

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$\rm\orange{Hope \ it \ helps \ :))}$