Q1
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) $x^2 – 2x – 8$
Factorization: Split middle term $-2x$ as $-4x + 2x$.
$$x^2 – 4x + 2x – 8 = x(x – 4) + 2(x – 4) = (x – 4)(x + 2)$$
For zeroes, put $p(x) = 0$: $x – 4 = 0$ or $x + 2 = 0$.
Zeroes are $\alpha = 4, \beta = -2$.
Verification: Here $a=1, b=-2, c=-8$.
$$\text{Sum: } \alpha + \beta = 4 + (-2) = 2 \quad \text{and} \quad -\frac{b}{a} = -\frac{(-2)}{1} = 2$$
$$\text{Product: } \alpha\beta = 4(-2) = -8 \quad \text{and} \quad \frac{c}{a} = \frac{-8}{1} = -8$$
✔️ Verified
(ii) $4s^2 – 4s + 1$
Factorization: Using identity $(a-b)^2$.
$$4s^2 – 4s + 1 = (2s – 1)^2$$
For zeroes: $2s – 1 = 0 \Rightarrow s = 1/2$.
Zeroes are $\alpha = 1/2, \beta = 1/2$.
Verification: Here $a=4, b=-4, c=1$.
$$\text{Sum: } \frac{1}{2} + \frac{1}{2} = 1 \quad \text{and} \quad -\frac{b}{a} = -\frac{(-4)}{4} = 1$$
$$\text{Product: } \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \quad \text{and} \quad \frac{c}{a} = \frac{1}{4}$$
✔️ Verified
(iii) $6x^2 – 3 – 7x$
Rearrange in standard form: $6x^2 – 7x – 3$.
Factorization: Split $-7x$ as $-9x + 2x$.
$$6x^2 – 9x + 2x – 3 = 3x(2x – 3) + 1(2x – 3) = (2x – 3)(3x + 1)$$
Zeroes: $2x-3=0 \Rightarrow x=3/2$ and $3x+1=0 \Rightarrow x=-1/3$.
Verification: $a=6, b=-7, c=-3$.
$$\text{Sum: } \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9-2}{6} = \frac{7}{6} \quad \text{and} \quad -\frac{b}{a} = -\frac{(-7)}{6} = \frac{7}{6}$$
$$\text{Product: } \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2} \quad \text{and} \quad \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$$
✔️ Verified
(iv) $4u^2 + 8u$
Factorization: Take $4u$ common.
$$4u^2 + 8u = 4u(u + 2)$$
Zeroes: $4u=0 \Rightarrow u=0$ and $u+2=0 \Rightarrow u=-2$.
Verification: $a=4, b=8, c=0$.
$$\text{Sum: } 0 + (-2) = -2 \quad \text{and} \quad -\frac{b}{a} = -\frac{8}{4} = -2$$
$$\text{Product: } 0 \times (-2) = 0 \quad \text{and} \quad \frac{c}{a} = \frac{0}{4} = 0$$
✔️ Verified
(v) $t^2 – 15$
Factorization: Use $a^2 – b^2 = (a-b)(a+b)$.
$$t^2 – 15 = t^2 – (\sqrt{15})^2 = (t – \sqrt{15})(t + \sqrt{15})$$
Zeroes: $t = \sqrt{15}$ and $t = -\sqrt{15}$.
Verification: $a=1, b=0, c=-15$.
$$\text{Sum: } \sqrt{15} – \sqrt{15} = 0 \quad \text{and} \quad -\frac{b}{a} = -\frac{0}{1} = 0$$
$$\text{Product: } \sqrt{15}(-\sqrt{15}) = -15 \quad \text{and} \quad \frac{c}{a} = \frac{-15}{1} = -15$$
✔️ Verified
(vi) $3x^2 – x – 4$
Factorization: Split $-x$ as $-4x + 3x$.
$$3x^2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)$$
Zeroes: $x = 4/3$ and $x = -1$.
Verification: $a=3, b=-1, c=-4$.
$$\text{Sum: } \frac{4}{3} + (-1) = \frac{1}{3} \quad \text{and} \quad -\frac{b}{a} = -\frac{(-1)}{3} = \frac{1}{3}$$
$$\text{Product: } \frac{4}{3}(-1) = -\frac{4}{3} \quad \text{and} \quad \frac{c}{a} = \frac{-4}{3}$$
✔️ Verified