Question

find the zeros of polynomial of the following polynomial and verify the relationship between its coefficient
x^2-49

Answer 1

$\huge\mathcal{Heya}$

$\mathsf{Given \ equation \ :-}$

$\mathsf{x^2 - 49}$

$\mathsf{On \ further \ solving, \ we \ get}$

$\mathsf{(x)^2 - (7)^2}$

$\mathsf{Using \ identity}$

$\mathsf{(a^2 - b^2) \ = \ (a + b)(a - b)}$

$\mathsf{Here,}$

$\mathsf{a \ = \ x \ And \ b \ = \ 7}$

$\mathsf{Therefore,}$

$\mathsf{= \ (x + 7)(x - 7)}$

To find, the zeroes, the two factors [(x + 7) and (x - 7)] should be equal to zero.

•°•

$\mathsf{(x + 7) \ = \ 0}$

$\mathsf{And}$

$\mathsf{(x - 7) \ = \ 0}$

•°•

$\mathsf{x \ = \ 7 \ and \ -7}$

$\mathsf{Zeroes \ are \ 7 \ and \ -7.}$

$\mathsf{Let}$

$\alpha = 7 \: and \: \beta = - 7$

$\huge\tt{Verification \ :-}$

All equations are in the form of :-

ax² + bx + c

So,

$\mathsf{Given \ equation \ will \ be}$

$\mathsf{x^2 + 0x - 49}$

$\mathsf{Here,}$

$\mathsf{a \ = \ 1, \ b \ = \ 0 \ and \ c \ = \ -49}$

$\mathsf{Now,}$

$\mathsf{Sum \ of \ zeroes \ :-}$

$= \alpha + \beta$
= $\mathsf{7 + (-7) \ = \ 0}$

$\mathsf{Also,}$

$\mathsf{Sum \ of \ zeroes \ :-}$

$= \frac{ - b}{a} \\ \\ = \frac{ - (0)}{1} = 0$

$\mathsf{Product \ of \ zeroes \ :-}$

$= \alpha \beta$
= $\mathsf{(7)(-7) \ = \ -49}$

$\mathsf{Also,}$

$\mathsf{Product \ of \ zeroes \ :-}$

$= \frac{c}{a} \\ \\ = \frac{ - 49}{1} = - 49$

$\huge\mathbb{Hope \ this \ helps.}$