Question

Find the zeros of the following quadratic polynomial and verify the Relationship between the zeros and the coefficient. 6x² - 7x - 3​

Answer 1

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Answer 2

Solution :

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

The quadratic polynomial is 6x² - 7x - 3.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The zeroes of the polynomial & verify the relationship between zeroes and coefficient.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

We have p(x) = 6x² - 7x - 3

Zero of the polynomial is p(x) = 0

So;

$\mapsto\sf{6x^{2} -7x-3=0}\\\\\mapsto\sf{6x^{2} -9x+2x-3=0}\\\\\mapsto\sf{3x(2x-3)+1(2x-3)=0}\\\\\mapsto\sf{(2x-3)(3x+1)=0}\\\\\mapsto\sf{2x-3=0\:\:\:Or\:\:\:3x+1=0}\\\\\mapsto\sf{2x=3\:\:\:Or\:\:\:3x=-1}\\\\\mapsto\sf{\pink{x=\dfrac{3}{2} \:\:Or\:\:\dfrac{-1}{3} }}$

∴ The α = 3/2 and β = -1/3 are the zeroes of the polynomial.

Now;

As the given quadratic polynomial as we compared with ax²+bx+c=0

• a = 6
• b = -7
• c = -3

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

$\mapsto\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coefficient\:of\:x^{2} }{Coefficient\:of\:x}}\\ \\\\\mapsto\sf{\dfrac{3}{2} +\bigg(\dfrac{-1}{3}\bigg) =\dfrac{-(-7)}{6} }\\\\\\\mapsto\sf{\dfrac{9+(-2)}{6} =\dfrac{7}{6}} \\\\\\\mapsto\sf{\dfrac{9-2}{6} =\dfrac{7}{6} }\\\\\\\mapsto\sf{\pink{\dfrac{7}{6} =\dfrac{7}{6} }}$

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

$\mapsto\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term }{Coefficient\:of\:x}}\\ \\\\\mapsto\sf{\dfrac{3}{2} \times \bigg(\dfrac{-1}{3}\bigg) =\dfrac{-3}{6} }\\\\\\\mapsto\sf{\pink{\dfrac{-3}{6} =\dfrac{-3}{6} }}$

Thus;

Relationship between zeroes and coefficient is verified .