Question

The sum of roots of equation x^2+px+q=0 is equal to sum of their squares, then prove that: p(p+1)=2q

Answer 1

Step-by-step explanation:

-p/1 = p^2 - 2q

-p = p^2 -2q

p^2+p= 2q

p(p+1) = 2q

Hence proved.

Answer 2

${x}^{2} + px + q = 0.....(i)$

$let \alpha , \beta \: be \: roots \: of \: eq(i)$

ACCORDING TO QUESTION

$\alpha + \beta = { \alpha }^{2} + { \beta }^{2} .....(ii)$

$now \: term \: eq(i) \alpha + \beta = - p$

$\alpha \beta = q$

$from \: eq(ii)$

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

$- p = ( - p {)}^{2} - 2q$

$- p = {p}^{2} - 2q$

${p}^{2} + p - 2q = 0$