Question

1. Find the zeroes of the polynomial f(x) = x2 – 9 and verify the relationship between the zeroes and the coefficients.

Answer 1

$\fbox{\texttt{\green{Answer:}}}$

Zeroes of the polynomial f(x) = 3 & -3

$\fbox{\texttt{\pink{Given:}}}$

Polynomial f(x) = x² - 9

$\fbox{\texttt{\blue{To\:find:}}}$

The zeroes of the polynomial f(x) and verify the relationship between zeroes and coefficients.

$\fbox{\texttt{\orange{Identity \:used:}}}$

a² - b² = (a + b)(a - b)

$\fbox{\texttt{\red{How\:to\:Find:}}}$

f(x) = x² - 9

Now,

$\implies$ x² - 9 = 0

$\implies$ x² - 3² = 0

$\implies$ (x + 3)(x - 3) = 0

$\implies$ x = $\bold{\pm{3}}$

$\therefore$ Zeroes of the quadratic polynomial f(x) = 3, -3

i.e., $\bold{\alpha=3\:and\:\beta=-3}$

$\rule{200}3$

★ Verification :

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial p(x) = ax² + bx + c, a $\bold{\neq}$ 0, then

• $\bold{\alpha+\beta=\dfrac{-b}{a}}$

• $\bold{\alpha\times{}\beta=\dfrac{c}{a}}$

Now,

$\bold{\alpha+\beta}$ = 3 + -3 = 0

$\bold{\alpha\times{}\beta}$ = 3 × -3 = -9

Therefore,

$\bold{\alpha+\beta=\dfrac{-b}{a}=\dfrac{0}{3}=0}$

$\bold{\alpha\times{}\beta=\dfrac{c}{a}=\dfrac{-9}{1}=\:-9}$

Hence, the relationship between zeroes and coefficients are verified.

Answer 2

Answer:

x^2-9

x^2+3x-3x-9

x(x+3) -3(x+3)

(x-3) (x+3)

x-3=0 , x+3=0

x=3,x=3

sum of the zeros=alpha+beta =3+(-3)

3-3=0

product of the zeros= alpha.beta = 3(-3)

3(-3) = -9

coefficient

-b/a=0/1 = 0

c/a=-9/1= -9