Question

alpha and beta are zeros of polynomial x^2-6x+k find the value of k such that ( alpha+beta)^2 -2 alpha.beta =40​

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

If $\alpha \: \: and \: \: \beta \:$are the zeroes of the quadratic polynomial $a {x}^{2} + bx + c$

Then

$\displaystyle \: \alpha + \beta \: = - \frac{b}{a} \: \: and \: \: \: \alpha \beta \: = \frac{c}{a}$

CALCULATION

Here $\alpha \: \: and \: \: \beta \:$are the zeroes of the quadratic polynomial ${x}^{2} - 6x + k$

Then

$\displaystyle \: \alpha + \beta \: = 6 \: \: and \: \: \: \alpha \beta \: = k$

${( \alpha + \beta )}^{2} - 2\alpha \beta = 40$

$\implies \: 36 - 2k = 40$

$\implies \: 2k = - 4$

So

$k = - 2$

Answer 2

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\sf\therefore \: p(x)=x^2-6x+k$

$\sf\dashrightarrow (\alpha+ \beta)^2-2 (\alpha . \beta)=40$

$\sf\dashrightarrow \alpha\:and\beta\:are\:zeroes\:of\:the\:polynomial$

$\large\underline\bold{TO\:FIND,}$

$\sf\dashrightarrow FIND \:THE\:VALUE\:OF\:'K'.$

$\bf\therefore\bold\red{NOTE:-\: \alpha \:and \:\beta\: are\: zeroes \:of \:the \:polynomials.\:then}$

$\sf\therefore (\alpha + \beta)= \dfrac{-b}{a}$

$\sf\therefore (\alpha . \beta)= \dfrac{c}{a}$

$\sf\therefore x^2-6x+k$

$\sf\dashrightarrow a=1$

$\sf\dashrightarrow b= -6$

$\sf\dashrightarrow c= k$

$\sf\dashrightarrow (\alpha + \beta)= \dfrac{-b}{a}$

$\sf\implies (\alpha + \beta)= \dfrac{-(-6)}{1}$

$\sf\implies (\alpha + \beta)= 6$

$\sf\dashrightarrow (\alpha . \beta)= \dfrac{c}{a}$

$\sf\implies (\alpha . \beta)= \dfrac{k}{1}$

$\sf\implies (\alpha . \beta)= k$

$\large\underline\bold{SOLUTION,}$

$\sf\therefore (\alpha+ \beta)^2-2 (\alpha . \beta)=40$

$\sf\therefore (\alpha + \beta)= 6$

$\sf\implies (\alpha . \beta)= k$

$\sf\therefore (6)^2-2 (k)=40$

$\sf\implies 36-2k=40$

$\sf\implies -2k=40-36$

$\sf\implies -2k= 4$

$\sf\implies -k= \dfrac{4}{2}$

$\sf\implies -k= \cancel \dfrac{4}{2}$

$\sf\implies -k=2$

$\sf\implies k=-2$

$\large{\boxed{\bf{\star\:\: k=-2\:\: \star }}}$

$\large\underline\bold{THE\:VALUE\:OF\:'k'\:\:IS\:-2}$

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