Question

Q-4) Find the zeroes of the quadratic polynomial 6x²+x-2​

Answer 1

Given :

• Polynomial 6x²+x-2.

To Find :

• Zeroes of the polynomial.

Solution :

$\longmapsto\tt\bold{{6x}^{2}+x-2}$

By Splitting Middle Term :

$\longmapsto\tt{{6x}^{2}+4x-3x-2}$

$\longmapsto\tt{2x(3x+2)-1(3x+2)}$

$\longmapsto\tt{(2x-1)(3x+2)}$

• $x=\dfrac{1}{2}$

• $x=\dfrac{-2}{3}$

So , 1/2 and -2/3 are the zeroes of polynomial 6x²+x-2...

_______________________

Here :

• a = 6
• b = 1
• c = -2

Sum of Zeroes :

$\longmapsto\tt{\dfrac{1}{2}+\dfrac{(-2)}{3}=\dfrac{-b}{a}}$

$\longmapsto\tt{\dfrac{1}{2}-\dfrac{2}{3}=\dfrac{-1}{6}}$

$\longmapsto\tt{\dfrac{3-4}{5}=\dfrac{-1}{6}}$

$\longmapsto\tt\bold{\dfrac{-1}{6}=\dfrac{-1}{6}}$

Product of Zeroes :

$\longmapsto\tt{\dfrac{1}{2}\times\dfrac{(-2)}{3}=\dfrac{c}{a}}$

$\longmapsto\tt{\cancel\dfrac{-2}{6}=\cancel\dfrac{-2}{6}}$

$\longmapsto\tt\bold{\dfrac{-1}{3}=\dfrac{-1}{3}}$

HENCE VERIFIED