Question

6
Polynomials
EXERCISE - 3.3
Find the zeroes of the following quadratic polynomials and verify the relationsh
the zeroes and the coefficients.
(1) x² - 2x - 8 ​

Answer 1

$\huge{\underline{\underline{\green{\mathfrak{Answer:}}}}}$

Finding zeroes

Given: P(x) = x² - 2x - 8

For zeroes of P(x)

=> P(x) = 0

=> x² - 2x - 8 = 0

By splitting method

=> x² + 2x - 4x - 8 = 0

=> x(x + 2) - 4(x + 2) = 0

=> (x - 4) (x + 2) = 0

=> x = 4, x = -2

Verification

we know that,

$\fbox{\mathbf{Sum\: of \:the \:zeroes\: = \frac{-(coefficient\: of \:x) }{coefficient\:of\:x^2}}}$

=> 4 + (-2) = $\mathbf{\frac{-(-2) }{1}}$

=> 2 = 2

{Both sides are equal}

$\fbox{\mathbf{Product \:of\: the\: zeroes \:= \frac{constant\:term }{coefficient\:of\:x^2}}}$

=> 4 × (-2) = $\mathbf{\frac{-8 }{1}}$

=> -8 = -8

{Both sides are equal}

Hence proved