Question

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeros and the coefficient 9x2-6x

Answer 1

Answer:

1/3

Step-by-step explanation:

9x²-6x = 0

b² - 4ac = 36-36 = 0

-b/2a = 6/18 = 1/3

Answer 2

Solution :(Ques.error)

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

The quadratic polynomial 9x² - 6x - 3.

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

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

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

We have p(x) = 9x² - 6x - 3

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\sf{9x^{2} -6x-3=0}\\\\\longrightarrow\sf{9x^{2} -9x+3x-3=0}\\\\\longrightarrow\sf{9x(x-1)+3(9x-1)=0}\\\\\longrightarrow\sf{(x-1)(9x+3)=0}\\\\\longrightarrow\sf{x-1=0\:\:\;Or\:\:\:9x+3=0}\\\\\longrightarrow\sf{x=1\:\:\:Or\:\:\:9x=-3}\\\\\longrightarrow\sf{x=1\:\:\:Or\:\:\:x=\cancel{\dfrac{-3}{9} }}\\\\\longrightarrow\sf{\orange{x=1\:\:Or\:\:x=-1/3}}$

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

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

;

• a = 9
• b = -6
• c = -3

Now;

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

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

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

$\longrightarrow\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coefficient\:of\:x^{2} } }\\\\\\\longrightarrow\sf{1\times \bigg(-\dfrac{1}{3} \bigg)=\dfrac{-3}{9} }\\\\\\\longrightarrow\sf{-\dfrac{1}{3} =\cancel{\dfrac{-3}{9}}}\\\\\\\longrightarrow\sf{\orange{-\dfrac{1}{3} =-\dfrac{1}{3} }}$

Thus;

Relationship between zeroes and coefficient is verified .