Question

find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients of x square - 36​

Answer 1

Solution :

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

The quadratic polynomial is x² - 36.

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

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

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

We have p(x) = x² - 36.

Zero of the polynomial p(x) = 0

So;

$\longrightarrow\sf{x^{2} -36=0}\\\\\longrightarrow\sf{x^{2} =36}\\\\\longrightarrow\sf{x=\pm\sqrt{36} }\\\\\longrightarrow\sf{\pink{x=\pm6}}$

∴ The α = 6 and β = -6 are the zeroes of the polynomial.  

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

• a = 1
• b = 0
• c = -36

Now;

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

$\longrightarrow\sf{\alpha +\beta =\dfrac{-b}{a} =\dfrac{Coeffiicient\:of\:x}{Coeffiicient\:of\:x^{2} } }\\\\\\\longrightarrow\sf{6+(-6)=\dfrac{0}{1} }\\\\\\\longrightarrow\sf{6-6=0}\\\\\\\longrightarrow\sf{\pink{0=0}}$

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

$\longrightarrow\sf{\alpha \times \beta =\dfrac{c}{a} =\dfrac{Constant\:term}{Coeffiicient\:of\:x^{2} } }\\\\\\\longrightarrow\sf{6\times (-6)=\dfrac{-36}{1} }\\\\\\\longrightarrow\sf{\pink{-36=-36}}$

Thus;

Relationship between zeroes and coefficient is verified .