Question

if zeroes of p(x) = 2x²-7x+k are reciprocal of ech find k ??​

Answer 1

Given

• p(x) = 2x² - 7x + k⠀⠀......[1]
• Zeroes of the polynomial are reciprocal of each other.

To find

• Value of k.

Solution

• Let the one zero be α.

$\sf\pink{⟶}$ We know that

$\underline{\boxed{\tt{Sum\: of\: zeroes = \dfrac{-b}{a}}}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {α + \dfrac{1}{α} = \dfrac{-(-7)}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {\dfrac{α^2 + 1}{α} = \dfrac{7}{2}}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {2(α^2 + 1) = 7α }$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {2α^2 + 2 = 7α}$

⠀

$\tt:\implies\: \: \: \: \: \: \: \: {2α^2 - 7α + 2 = 0}$⠀⠀.....[2]

⠀

$\sf\pink{⟶}$ On comparing equation [1] and [2], we get

$\large{\underline{\boxed{\bf{\orange{k = 2}}}}}$

━━━━━━━━━━━━━━━━━━━━━━

Answer 2

$\huge{\underline{\underline{Given→}}}$

• $\sf\green{Equation→p(x)=2x^2-7x+k}$
• $\sf\green{The\:roots\:are\:reciprocal\:of\:each\:other}$

$\huge{\underline{\underline{To\:Find→}}}$

$\sf\purple{→The\:value\:of\:k.}$

$\huge{\underline{\underline{Answer→}}}$

So let the roots be $α\:and\:β$

And the roots are reciprocal,so $β=\dfrac{1}{α}$

Relations of Zeroes & Coefficients:-

• $\bold{\underline{\underline{Sum\:of\:Zeroes\dashrightarrow}}}$

$\sf\blue{→α+β=\dfrac{-b}{a}}$

$\sf\blue{→α+\dfrac{1}{α}=\dfrac{-b}{a}}$

$\sf\blue{→\dfrac{α^2+1}{α}=\dfrac{-b}{a}}$

$\sf\blue{→\dfrac{α^2+1}{α}=\dfrac{-(-7)}{2}}$

$\sf\blue{→\dfrac{α^2+1}{α}=\dfrac{7}{2}......(i)}$

• $\bold{\underline{\underline{Product\:of\:zeroes\dashrightarrow}}}$

$\sf\blue{→α\timesβ=\dfrac{c}{a}}$

$\sf\blue{→α\times\dfrac{1}{α}=\dfrac{c}{a}}$

$\sf\blue{→1=\dfrac{c}{a}}$

$\sf\blue{→1=\dfrac{k}{2}......(ii)}$

From eq. (ii):-

$\sf\blue{→1=\dfrac{k}{2}}$

$\sf\blue{→k=1\times2}$

$\sf\blue{→k=2}$

$\sf{\boxed{\boxed{\pink{→k=2✔}}}}$

☞ Hence the value of k(constant term) is 2 which is the required answer.

HOPE IT HELPS.