Question

3 Find the zeroes of the 6x - 3-78 quadratic polynomials and verify the relationship between the and the co-efficients. Zeroes​

Answer 1

Correct Question :-

find the zeroes of the quadratic polynomial 6x square - 3 - 7x and verify the relation between the zero and the coefficient of the polynomial

To Find :-

Zeroes

Solution :-

$\sf 6x^2-3-7x=0$

$\sf 6x^2-7x-3=0$

$\sf 6x^2-(9x-2x)-3=0$

$\sf 6x^2-9x+2x-3=0$

$\sf 3x(2x-3)+1(2x-3)=0$

$\sf (2x-3)(3x+1)=0$

$\sf x=\dfrac{3}{2}\;\&\;\dfrac{-1}{3}$

V e r i f i c a t i o n :

On comparing the given equation with ax² + bx + c

${\large{\boxed{\underline{\underline{\bf \dag\alpha +\beta =\dfrac{-b}{a}}}}}}$

$\sf \dfrac{3}{2}+\dfrac{-1}{3}=\dfrac{-(-7)}{6}$

$\sf \dfrac{9+(-2)}{6}=\dfrac{7}{6}$

$\sf\dfrac{7}{6}=\dfrac{7}{6}$

${\large{\boxed{\underline{\underline{\bf \dag\alpha \beta =\dfrac{c}{a}}}}}}$

$\sf\dfrac{3}{2}\times\dfrac{-1}{3}=\dfrac{-3}{6}$

$\sf\dfrac{-3}{6}=\dfrac{-1}{2}$

$\sf\dfrac{-1}{2}=\dfrac{-1}2$

Answer 2

$6 {x}^{2} - 3 - 7 x= 0$

$6 {x}^{2} - 7x - 3 = 0$

$6 {x}^{2} - (9 - 2)x - 3 = 0$

$6 {x}^{2} - 9x + 2x - 3 = 0$

$2x(3x + 1) - 3(3x + 1) = 0$

$(3x + 1)(2x - 3) = 0$

so , zeros are -⅓ and , 3/2