Question

Find the sum and product of zeros of the polynomial 6x²-3-7x

Answer 1

Given :

• Polynomial 6x²-3-7x

To Find :

• Sum and Product of Zeroes

Solution :

$\implies\sf\bold{{6x}^{2}-7x-3}$

By Splitting Middle Term :-

$\implies\sf{{6x}^{2}-(9x-2x)-3}$

$\implies\sf{{6x}^{2}-9x+2x-3}$

$\implies\sf{3x(2x-3)+1(2x-3)}$

$\implies\sf{(3x+1)(2x-3)}$

• x = $\dfrac{-1}{3}$

• x = $\dfrac{3}{2}$

So , -1/3 and 3/2 are the zeroes of polynomial 6x²-7x-3...

_____________________

Here :

• a = 6
• b = -7
• c = -3

Sum of Zeroes :

$\implies\sf{\alpha+\beta=\dfrac{-b}{a}}$

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

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

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

$\implies\sf\bold{L.H.S=R.H.S}$

Product of Zeroes :

$\implies\sf{\alpha\beta=\dfrac{c}{a}}$

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

$\implies\sf{\cancel\dfrac{-3}{6}=\cancel\dfrac{-3}{6}}$

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

$\implies\sf\bold{L.H.S=R.H.S}$