# Find the zeros of the following quadratic polynomials and verify

Question.

Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and the coefficients.

(i) $x^{2}-2 x-8$

(ii) $4 \mathrm{~s}^{2}-4 \mathrm{~s}+1$

(iii) $6 x^{2}-3-7 x$

(iv) $4 u^{2}+8 u$

(v) $t^{2}-15$

(vi) $3 x^{2}-x-4$

Solution:

(i) $x^{2}-2 x-8=x^{2}-4 x+2 x-8$

$=x(x-4)+2(x-4)=(x+2)(x-4)$

Zeroes are – 2 and 4.

Sum of the zeros

$=(-2)+(4)=2=\frac{-(-2)}{1}=\frac{-(\text { Coefficient of } \mathbf{x})}{\left(\text { Coefficient of } \mathbf{x}^{\mathbf{2}}\right)}$

Product of the zeros

$=(-2)(4)=-8=\frac{(-8)}{1}=\frac{(\text { Constant temi) }}{\left(\text { Coefficient of } \mathbf{x}^{2}\right)}$

(ii) $4 \mathrm{~s}^{2}-4 \mathrm{~s}+1=(2 \mathrm{~s}-1)^{2}$

The two zeros are $\frac{\mathbf{1}}{\mathbf{z}}, \frac{\mathbf{1}}{\mathbf{2}}$

Sum of the two zeros

$=\frac{1}{2}+\frac{1}{2}=1=\frac{-(-4)}{4}=\frac{-(\text { Coeficient of } x)}{\left(\text { Coeficient of } x^{2}\right)}$

Product of two zeros

$=\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)=\frac{1}{4}=\frac{(\text { Constant temm })}{\left(\text { Coeficient of } x^{2}\right)}$

(iii) $6 x^{2}-7 x-3$

$=6 x^{2}-9 x+2 x-3$

$=3 x(2 x-3)+1(2 x-3)$

$=(2 x-3)(3 x+1)$

zeros are $\frac{\mathbf{3}}{\mathbf{2}}, \frac{\mathbf{1}}{\mathbf{3}}$

Sum of zeros $=\frac{\mathbf{3}}{\mathbf{2}}+\left(\frac{-\mathbf{1}}{\mathbf{3}}\right)$

$=\frac{\mathbf{9}-\mathbf{2}}{\mathbf{6}}=\frac{\mathbf{7}}{\mathbf{6}}=\frac{-(\mathbf{7})}{\mathbf{6}}=\frac{-(\text { coefficient of } \mathbf{x})}{\left(\text { coefficient of } \mathbf{x}^{2}\right)}$

Product of zeros

$=\frac{3}{2} \times\left(\frac{-1}{3}\right)=\frac{-1}{2}=\frac{-(3)}{6}$

$=\frac{(\text { constant temm })}{\left(\text { coefficient of } x^{2}\right)}$

(iv) $4 \mathrm{u}^{2}+8 \mathrm{u}=4 \mathrm{u}(\mathrm{u}+2)$

zeros are $0,-2$

Sum of zeros

$=0+(-2)=-2=\frac{-(\mathbf{8})}{\mathbf{4}}$

$=-\frac{(\text { coefficient of u) }}{\left(\text { coeficient of ư }^{2}\right)}$

Product of zeros

$=0 \times(-2)=0=\frac{\mathbf{0}}{\mathbf{4}}$

$=\frac{\text { comstant temm }}{\text { cosficsert of ne }^{2}}$

(v) $\mathrm{t}^{2}-15=(\mathrm{t}-\sqrt{\mathbf{1 5}})(\mathrm{t}+\sqrt{\mathbf{1 5}})$

zeros are $\sqrt{\mathbf{1 5}},-\sqrt{\mathbf{1 5}}$

sum of zeros

$=\sqrt{15}+(-\sqrt{15})=0=\frac{0}{1}$

$=-\frac{(\text { coefficient of } t)}{\left(\text { coefficient of } t^{2}\right)}$

Product of zeros

$=(\sqrt{15})(-\sqrt{15})=-15=\frac{-15}{1}$

$=\frac{\text { constant tem }}{\text { coefficient of } t^{2}}$

(vi) $3 x^{2}-x-4$

$=3 x^{2}-4 x+3 x-4$

$=x(3 x-4)+1(3 x-4)$

$=(3 x-4)(x+1)$

zeros are $\frac{4}{3},-1$

Sum of zeros

$=\frac{4}{3}-1=\frac{1}{3}=-\frac{(-1)}{3}$

$=-\frac{(\text { coeficient of } x)}{\text { coeficient of } x^{2}}$

Product of zeros $=\frac{4}{3} \times(-1)=-\frac{4}{3}$

$=\frac{(\text { constant temn })}{\text { coeficient of } x^{2}}$