Question

if the zeroes of the polynomial x^3-9x^2+2x+21 are p-q, p, p+q, find the values of p and q​

Answer 1

$\underline{\textbf{Given:}}$

$\mathsf{p-q,p,p+q\;are\;zerroes\;of\;x^3-9x^2+2x+21}$

$\underline{\textbf{To find:}}$

$\textsf{The values of p and q}$

$\underline{\textbf{Solution:}}$

$\mathsf{Consider,\;\;x^3-9x^2+2x+21}$

$\mathsf{Sum\;of\;the\;roots=\dfrac{-(-9)}{1}}$

$\implies\mathsf{p-q+p+p+q=9}$

$\implies\mathsf{3p=9}$

$\implies\mathsf{p=\dfrac{9}{3}}$

$\implies\boxed{\mathsf{p=3}}$

$\mathsf{Product\;of\;the\;roots=\dfrac{-(21)}{1}}$

$\implies\mathsf{(p-q)p(p+q)=-21}$

$\implies\mathsf{(p^2-q^2)p=-21}$

$\implies\mathsf{(3^2-q^2)3=-21}$

$\implies\mathsf{9-q^2=\dfrac{-21}{3}}$

$\implies\mathsf{9-q^2=-7}$

$\implies\mathsf{-q^2=-7-9}$

$\implies\mathsf{-q^2=-16}$

$\implies\mathsf{q^2=16}$

$\implies\boxed{\mathsf{q=\pm\,4}}$

$\therefore\mathsf{p=3\;\;\&\;\;q=\pm\,4}$

$\underline{\textbf{Find more:}}$

If the zeroes of cubic polynomial x3-15x2+74x-120 are the form a,a+b and a+2b for some positive real number, than the value of a and b are​

https://brainly.in/question/40663788

Answer 2

Given is a polynomial Equation: $x^3-9x^2+2x+21$=0

And its zeroes are p-q ,p & p+q

We need to find the value of p and q

As we know the sum of zeroes is =  $\frac{-b}{a}$ and the product the roots gives = $\frac{-d}{a}$

Here in the given equation:-

$ax^3-bx^2+cx+d$ $=x^3-9x^2+2x+21$

i.e. $a=1, b=9,c=2 , d= 21$

Hence, the sum of zeroes = $\frac{-b}{a}$

           $(p+q)+p+(p-q)=\frac{-(-9)}{1}$

                   $3p=\frac{-(-9)}{1}$

                    $p=\frac{-(-9)}{3}$    

                    $p=\frac{9}{3}=3$        

∴p=3    

Also the product of zeroes is = $\frac{-d}{a}$

  $(p+q)*p*(p-q)=\frac{-21}{1}$

                $p(p^{2} -q^{2}) =\frac{-21}{1}$

                   $p^{3} -pq^{2} =\frac{-21}{1}$

                   $3^{3} -3q^{2} =\frac{-21}{1}$

                        $-3q^{2} =-21-3^{3}$

                        $-3q^{2} =-21-27$    

                        $-3q^{2} =-48$    

                            $q^{2} =\frac{48}{3}$

                            $q^{2} =16$

                             $q=\sqrt[2]{16} =4$

∴q$=4$

So, the value of p=3 and q$=4$