Question

1. Find the zeroes of the following quadratic polynomials and verify the relationship between
the zeroes and the coefficients.
3x2-x-4
​

Answer 1

$\large{\textsf{Answer :-}}$

3x² - x - 4

3x² - 4x + 3x - 4

x (3x - 4) + 1 (3x - 4)

(3x - 4) (x + 1)

$\implies$ 3x² - x - 4 = 0

$\implies$ (3x - 4) (x + 1) = 0

$\implies$ (3x - 4) = 0

3x = 4

x = $\dfrac{4}{3}$

$\implies$ (x + 1) = 0

x = -1

So, $\dfrac{4}{3}$ & -1 are zeroes of polynomial 3x² - x - 4.

$\large{\textsf{Verification :-}}$

Sum of zeroes = $\dfrac{-b}{a}$

$\dfrac{4}{3}$ + (-1) = $\dfrac{-(-1)}{3}$

$\dfrac{4 - 3}{3}$ = $\dfrac{1}{3}$

$\dfrac{1}{3}$ = $\dfrac{1}{3}$

Also, Product of zeroes = $\dfrac{c}{a}$

$\dfrac{4}{3}$ × -1 = $\dfrac{-4}{3}$

$\dfrac{-4}{3}$ = $\dfrac{-4}{3}$

$\underline{\text{Hence, Verified..!}}$

Answer 2

Solution:-

Given Quadratic Equation:-

:=)3x² - x - 4 = 0

:=) 3x² - ( 4 -3)x - 4 = 0

:=) 3x² - 4x + 3x - 4 = 0

:=) x ( 3x - 4) + 1( 3x - 4) = 0

:=) ( 3x - 4) ( x + 1) = 0

:=) 3x = 4. and x = -1

:=) x = 4/3. and x = -1.

Hence,

Zeroes of the Given Quadratic Equation are (-1) and ( 4/3).

Verification:-

Case |,

Sum of Zeroes = -b/a

:=) 4/3 + (-1) = -(-1)/3

:=) ( 4 - 3)/3 = 1/3

:=) 1/3 = 1/3

Case ||,

Product of Zeroes = c/a

:=) 4/3 × (-1) = -4/3

:=) -4/3 = -4/3

Hence Solved!