Question

If one of the zeroes of the quadratic polynomial x² + px + 1 is - 3, then find the value of p ??​

Answer 1

Answer:

11/34 is the answer of this question

Answer 2

ANSWER✔

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

$\sf\dashrightarrow p(x)= x^2+px+1$

$\sf\dashrightarrow one\:of\:the\:zeroes\:the\:polynomial\:is,\:-3$

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

$\sf\large\dashrightarrow THR\:VALUE\:OF\:'p'.$

✯.PROPERTY USED,

$\sf {\boxed{\bf{\:\: to\:find\:zeroes\:of\:the\:polynomial\:we\:need\:p(x)=0}}}$

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

$\sf\therefore p(x)= x^2+px+1$

$\sf\therefore x=-3$

$\sf\dashrightarrow x^2+px+1=0\:-----\big[ by\:given\:property. \big]$

✯PUTTING THE VALUE OF 'x' IN THE GUVEN EQUATION. WE GET,

$\sf\implies x^2+px+1=0$

$\sf\implies (-3)^2+p(-3)+1=0$

$\sf\implies (9)+(-3p)+1=0$

$\sf\implies 10+(-3p)=0$

$\sf\implies -3p=-10$

$\sf\implies \cancel{-} \: 3p = \cancel{-} \: 10$

$\sf\implies 3p=10$

$\sf\implies p= \dfrac{10}{3}$

$\large{\boxed{\bf{\star\:\: p= \dfrac{10}{3}\:\: \star }}}$

✯VERIFICATION,

$\sf\therefore p(x)= x^2+px+1$

$\sf\therefore x=-3$

$\sf\therefore p= \dfrac{10}{3}$

$\sf\implies (-3)^2+\bigg(\dfrac{10}{3} \bigg)(-3)+1=0$

$\sf\implies (9)+ \bigg(\dfrac{10}{\cancel{3}} \bigg)(\cancel{-3}) +1=0$

$\sf\implies 9+ (-10)+1=0$

$\sf\implies (-1)+1=0$

$\sf\implies 0=0$

$\sf\therefore L.H.S=R.H.S$

HENCE VERIFYED✔

$\large\underline\bold{\therefore \: THE \:VALUE \: OF\:'p'\:IS \dfrac{10}{3}}$

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