Question

find the Zeroes of 6x²-72-3 and verify the relation-
-shie between zeroes
and coefficients.​

Answer 1

S O L U T I O N : (Ques.error)

We have quadratic polynomial p(x) = 6x² - 7x - 3 & zeroes of the polynomial p(x) = 0

$\underline{\tt{Using\:by\:factorization \:\:method\::}}$

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

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

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

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

$\mapsto\sf{2x - 3 = 0\:\:\:Or\:\:\:3x + 1 = 0}$

$\mapsto\sf{2x =3\:\:\:Or\:\:\:3x =-1}$

$\mapsto\sf{x =3/2\:\:\:Or\:\:\:x=-1/3}$

∴ α = 3/2 & β = -1/3 are two zeroes of the polynomials .

As we know that quadratic polynomial compared with ax² + bx + c;

• a = 6
• b = -7
• c = -3

Now,

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\longrightarrow\tt{\alpha + \beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\longrightarrow\tt{\dfrac{3}{2} +\bigg(- \dfrac{1}{3} \bigg) = \dfrac{-(-7)}{6}}$

$\longrightarrow\tt{\dfrac{3}{2} - \dfrac{1}{3} = \dfrac{7}{6}}$

$\longrightarrow\tt{\dfrac{9-2}{6} = \dfrac{7}{6}}$

$\longrightarrow\bf{\dfrac{7}{6} = \dfrac{7}{6}}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\longrightarrow\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\longrightarrow\tt{\dfrac{3}{2} \times \bigg(- \dfrac{1}{3} \bigg) = \dfrac{-3}{6}}$

$\longrightarrow\bf{\dfrac{-3}{6} = \dfrac{-3}{6}}$

Thus,

The relationship between zeroes & coefficient are verified .