Question

if alpha and beta are the zeroes of polynomial p(x)= x^2-px+q. find the value of alpha square+beta square​

Answer 1

Answer:

p^2 - 2q

Step-by-step explanation:

Polynomials written in form of x^2 - Sx + P = 0 represent S as sum of their roots and P as product of roots. So, here if α and β are roots.

  sum of roots = α + β = p

  product of roots = αβ = q

⇒ α + β = p

⇒ ( α + β )^2 = ( p )^2

⇒ α^2 + β^2 + 2αβ = p^2

⇒ α^2 + β^2 + 2( q ) = p^2  { αβ = q }

⇒ α^2 + β^2 + 2q = p^2

⇒ α^2 + β^2 = p^2 -2q

Answer 2

Answer:

Given : $\alpha$ and $\beta$ are the zeroes of the polynomial $\rm p(x)=x^2-px+q$.

$\rule{80}{0.8}$

$\underline{\bigstar\:\textsf{We Have to find the value of :}}$

$:\implies\sf \bigg(\alpha \bigg)^2+ \bigg(\beta\bigg)^2\\\\\\:\implies\sf \bigg(\alpha + \beta \bigg)^2 - 2\alpha \beta \\\\\\{\sf\qquad\because\:(a+b)^2=a^2+b^2+2ab}\\\\\\:\implies\sf\bigg(Sum\:of\: Zeroes\bigg)^2 -\bigg(2Product\:of\: Zeroes\bigg)\\\\\\:\implies\sf \bigg(\dfrac{-\:b}{a}\bigg)^2 - \bigg(2 \times \dfrac{c}{a}\bigg)\\\\\\{\sf\qquad\because\:a=1\quad b=-\:p\quad c=q}\\\\\\:\implies\sf \bigg(\dfrac{-\:(- p)}{1}\bigg)^2 - \bigg(2 \times \dfrac{q}{1}\bigg)\\\\\\:\implies\large\underline{\boxed{\sf p^2 - 2q}}$

⠀

$\therefore\:\underline{\textsf{Hence, required value of the given is \textbf{(p ^\text2 - 2q)}}}.$

$\rule{180}{1.5}$

$\underline{\bigstar\:\textsf{Standard Form of Quadratic Polynomial :}}$

$\dashrightarrow\sf\:\:x^2-(Sum\:of\: Zeroes)x+(Product\: of\: Zeroes)=0\\\\\\\dashrightarrow\sf\:\:x^2-(\alpha+\beta)x+(\alpha\beta)=0$