Exercise-2.2, Class 10th, Maths, Chapter 2, NCERT

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

Recall: For quadratic $a{x}^2+bx+c$ with zeroes $\alpha ,\beta$

$$\alpha +\beta =-\frac{b}{a},\alpha \beta =\frac{c}{a}\mathrm{.}$$

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

Factor: $x^2-2x-8=(x-4)(x+2)$

Zeroes: $4$ and $-2$.

Sum $=4+(-2)=2=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{-2}{1}}=2.$

Product $=4\times (-2)=-8={\displaystyle \frac{c}{a}}={\displaystyle \frac{-8}{1}}\mathrm{.}$

---

(ii) $4s^2-4s+1$

Recognize: $(2s-1{)}^2=4s^2-4s+1$.

Zero (double): $s={\displaystyle \frac{1}{2}}$ (repeated root).

Sum $={\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2}}=1=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{-4}{4}}=1$

Product $={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}={\displaystyle \frac{c}{a}}={\displaystyle \frac{1}{4}}\mathrm{.}$

---

(iii) $6x^2-7x-3$
$\mathrm{=\; 6}{x}^2-3-7x$

Compute discriminant: $D=(-7{)}^2-4\cdot 6\cdot (-3)=49+72=121$, $\sqrt{D}=11$

Roots:

$$x=\frac{7\pm 11}{12}\Rightarrow x=\frac{18}{12}=\frac{3}{2},x=\frac{-4}{12}=-\frac{1}{3}\mathrm{.}$$

Sum $=\frac{3}{2}-\frac{1}{3}=\frac{9-2}{6}=\frac{7}{6}=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{-7}{6}}=\frac{7}{6}\mathrm{.}$

Product $=\frac{3}{2}\cdot (-\frac{1}{3})=-\frac{1}{2}={\displaystyle \frac{c}{a}}={\displaystyle \frac{-3}{6}}=-\frac{1}{2}\mathrm{.}$

---

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

Factor: $4u^2+8u=4u(u+2)$.

Zeroes: $u=0$ and $u=-2$.

Sum $=0+(-2)=-2=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{8}{4}}=-2.$
Product $=0\cdot (-2)=0={\displaystyle \frac{c}{a}}={\displaystyle \frac{0}{4}}\mathrm{.}$

---

(v) $t^2-15$

Use difference of squares: $t^2-15=(t-\sqrt{15})(t+\sqrt{15})$

Zeroes: $\sqrt{15}$ and $-\sqrt{15}$.

Sum $=0=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{0}{1}}\mathrm{.}$
Product $=-(\sqrt{15}{)}^2=-15={\displaystyle \frac{c}{a}}\mathrm{.}$

---

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

Discriminant: $D=(-1{)}^2-4\cdot 3\cdot (-4)=1+48=49,\sqrt{D}=7.$

Roots:

$$x=\frac{1\pm 7}{6}\Rightarrow x=\frac{8}{6}=\frac{4}{3}\text{ ? — check: }\frac{8}{6}=\frac{4}{3}\text{and}x=\frac{-6}{6}=-1.$$

Zeroes: ${\displaystyle \frac{4}{3}}$ and $-1$.

Sum $=\frac{4}{3}+(-1)=\frac{1}{3}=-{\displaystyle \frac{b}{a}}=-{\displaystyle \frac{-1}{3}}=\frac{1}{3}\mathrm{.}$
Product $={\displaystyle \frac{4}{3}}\times (-1)=-{\displaystyle \frac{4}{3}}={\displaystyle \frac{c}{a}}={\displaystyle \frac{-4}{3}}\mathrm{.}$