Question

Find the zeroes and verify the relationship between the zeroes and their coefficients.
$\left(i\right)3x^2+4x-4$
$\left(ii\right)21t^2-t-2$
$\left(iii\right)2u^2-7$

Answer 1

$i).3x^2+4x-4$

$\Rightarrow 3x^2+6x-2x-4$

$3x(x+2)-2(x+2)$

$(3x-2)(x+2)$

$\Rightarrow$ Zeroes : $x=\frac{2}{3},-2$

Sum of zeroes $=\frac{2}{3}+(-2)=\frac{-4}{3}=\frac{-b}{a}$

Product of zeroes $=\left(\frac{2}{3}\right)(-2)=\frac{-4}{3}=\frac{c}{a}$

The sum of zeroes and product of zeroes are equal to $\frac{-b}{a}\&\frac{c}{a}$ respectively.

$ii).21t^2-t-2$

$\Rightarrow 21t^2-7t+6t-2$

$7t(3t-1)+2(3t-1)$

$(7t+2)(3t-1)$

$\Rightarrow$ Zeroes $:t=\frac{-2}{7}\&\frac{1}{3}$

Sum of zeroes $=\frac{-2}{7}+\frac{1}{3}=\frac{1}{21}=\frac{-b}{a}$

Product of zeroes $=\left(\frac{-2}{7}\right)\left(\frac{-2}{7}\right)=\left(\frac{1}{3}\right)=\frac{c}{a}$ verified.

$iii).2u^2-7$

$\Rightarrow u=\sqrt{\frac{7}{2}},-\sqrt{\frac{7}{2}}$

$\Rightarrow$ Sum of zeroes $=0=\frac{-b}{a}$

Product of zeroes $=\frac{-7}{2}=\frac{c}{a}$ verified.

Hence, solved.