Question

Find the zeros of the following quadratic
polynomials and verify the relationship between
and their coefficients
f (y) = 7y²-11/3-2/3​

Answer 1

Solution :

We have quadratic polynomial f(y) = 7y² - 11/3y - 2/3

Zero of the polynomial f(y) = 0

$\longrightarrow\sf{7y^{2} - \dfrac{11}{3} y - \dfrac{2}{3} =0}\\\\\longrightarrow\sf{21y^{2} - 11y - 2 = 0}\\\\\longrightarrow\sf{21y^{2} -14y + 3y-2 = 0}\\\\\longrightarrow\sf{7y(3y-2) + 1 (3y -2) = 0}\\\\\longrightarrow\sf{(3y-2)(7y+1)=0}\\\\\longrightarrow\sf{3y-2=0\:\:\:Or\:\:\:7y+1=0}\\\\\longrightarrow\sf{3y=2\:\:\:Or\:\:\:7y=-1}\\\\\longrightarrow\bf{y=2/3\:\:\:Or\:\:\:y=-1/7}$

∴ We get α = 2/3 & β = -1/7 are the zeroes of the polynomial.

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

• a = 21
• b = -11
• c = 2

Now;

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

$\longrightarrow\sf{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\longrightarrow\sf{\dfrac{2}{3}+\bigg(-\dfrac{1}{7} \bigg)=\dfrac{-(-11)}{21}} \\\\\\\longrightarrow\sf{\dfrac{2}{3}-\dfrac{1}{7} =\dfrac{11}{21}} \\\\\\\longrightarrow\sf{\dfrac{14-3}{21} =\dfrac{11}{21}} \\\\\\\longrightarrow\bf{\dfrac{11}{21} =\dfrac{11}{21} }$

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

$\longrightarrow\sf{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg\rgroup}\\\\\\\longrightarrow\sf{\dfrac{2}{3}\times \bigg(-\dfrac{1}{7} \bigg)=\dfrac{-2}{21}} \\\\\\\longrightarrow\bf{\dfrac{-2}{21} =\dfrac{-2}{21} }$

Thus;

The relationship between zeroes & coefficient is verified .