Question

Find the zeros in the following quadratic polynomial. Also verify the relationship between the zeroes and its coefficients :

$3x^2-5x-2$

$9y^2-6y+1$

$4z^2+12z+9$

$3x^2-x$

Answer 1

Here multiple options are correct.

$3x^2-5x-2$

and

$3x^2-x$

are correct.

A

$3x^2-5x-2=0$

$\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\frac{5\pm \sqrt{25+24}}{6}$

$\frac{5\pm 7}{6}$

$\alpha =2,\beta =\frac{-1}{3}$

$\alpha +\beta =\frac{-b}{a}$

$2\frac{-1}{3}=\frac{5}{3}$

$\frac{5}{3}=\frac{5}{3}$

B

$9y^2-6y+1=0$

$9y^2-3y-3y+1=0$

$3y(3y-1)-1(3y-1)=0$

$(3y-1)(3y-1)=0$

$y=\frac{1}{3}$

$\alpha =\frac{1}{3},\beta =\frac{1}{3}$

$\alpha +\beta =\frac{-b}{a}$

$\frac{1}{3}+\frac{1}{3}=\frac{2}{3}$

$\alpha \beta =\frac{1}{3}\frac{1}{3}=\frac{1}{9}$

C

$4z^2+12z+9=0$

$4z^2+6z+6z=9=0$

$2z(2z+3)+3(2z+3)=0$

$(2z+3)(2z+3)=0$

$y=\frac{-3}{2}$

$\alpha =\frac{-3}{2},\beta =\frac{-3}{2}$

$\alpha +\beta =\frac{-b}{a}$

$\frac{-3}{2}+\frac{-3}{2}=\frac{-12}{4}$

$-3=-3$

Similarly $\alpha \beta =\frac{c}{a}=\frac{9}{4}$

D

$3x^2-x=0$

$x(3x-1)=0$

$x=0,\frac{1}{3}$

$\alpha =0,\beta =\frac{1}{3}$

$\alpha +\beta =\frac{-b}{a}$

$0+\frac{1}{3}=\frac{1}{3}$

$\frac{1}{3}=\frac{1}{3}$

$\alpha \beta =\frac{c}{a}$=0