Question 1
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients:
(i) $x^2 - 5x + 6$
(ii) $9x^2 - 12x + 4$
(iii) $6x^2 - 5x - 6$
(iv) $4x^2 + 12x$ (v) $t^2 - 20$ (vi) $3x^2 - 5x - 2$
(iv) $4x^2 + 12x$ (v) $t^2 - 20$ (vi) $3x^2 - 5x - 2$
(i) $x^2 - 5x + 6$
Factorise by splitting the middle term:
$$x^2 - 5x + 6 = x^2 - 2x - 3x + 6 = (x - 2)(x - 3)$$
To find zeroes, set $p(x) = 0 \implies x = 2 \text{ or } x = 3$.
So, $\alpha = 2$ and $\beta = 3$.
Here, $a = 1, b = -5, c = 6$.
Verification:
Sum of zeroes: $\alpha + \beta = 2 + 3 = 5 = -\dfrac{-5}{1} = -\dfrac{b}{a} \quad \checkmark$
Product of zeroes: $\alpha\beta = 2 \times 3 = 6 = \dfrac{6}{1} = \dfrac{c}{a} \quad \checkmark$
So, $\alpha = 2$ and $\beta = 3$.
Here, $a = 1, b = -5, c = 6$.
Verification:
Sum of zeroes: $\alpha + \beta = 2 + 3 = 5 = -\dfrac{-5}{1} = -\dfrac{b}{a} \quad \checkmark$
Product of zeroes: $\alpha\beta = 2 \times 3 = 6 = \dfrac{6}{1} = \dfrac{c}{a} \quad \checkmark$
(ii) $9x^2 - 12x + 4$
Factorise using identity $(a - b)^2 = a^2 - 2ab + b^2$:
$$9x^2 - 12x + 4 = (3x - 2)^2$$
Setting $p(x) = 0 \implies 3x - 2 = 0 \implies x = \dfrac{2}{3}, \dfrac{2}{3}$.
So, $\alpha = \dfrac{2}{3}, \beta = \dfrac{2}{3}$.
Here, $a = 9, b = -12, c = 4$.
Verification:
Sum of zeroes: $\alpha + \beta = \dfrac{2}{3} + \dfrac{2}{3} = \dfrac{4}{3} = -\dfrac{-12}{9} = -\dfrac{b}{a} \quad \checkmark$
Product of zeroes: $\alpha\beta = \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9} = \dfrac{c}{a} \quad \checkmark$
So, $\alpha = \dfrac{2}{3}, \beta = \dfrac{2}{3}$.
Here, $a = 9, b = -12, c = 4$.
Verification:
Sum of zeroes: $\alpha + \beta = \dfrac{2}{3} + \dfrac{2}{3} = \dfrac{4}{3} = -\dfrac{-12}{9} = -\dfrac{b}{a} \quad \checkmark$
Product of zeroes: $\alpha\beta = \dfrac{2}{3} \times \dfrac{2}{3} = \dfrac{4}{9} = \dfrac{c}{a} \quad \checkmark$
(iii) $6x^2 - 5x - 6$
Factorise: product = $-36$, sum = $-5 \implies -9$ and $4$:
$$6x^2 - 9x + 4x - 6 = 3x(2x - 3) + 2(2x - 3) = (3x + 2)(2x - 3)$$
Setting $p(x) = 0 \implies x = -\dfrac{2}{3}$ or $x = \dfrac{3}{2}$.
So, $\alpha = -\dfrac{2}{3}, \beta = \dfrac{3}{2}$.
Here, $a = 6, b = -5, c = -6$.
Verification:
Sum: $\alpha + \beta = -\dfrac{2}{3} + \dfrac{3}{2} = \dfrac{-4 + 9}{6} = \dfrac{5}{6} = -\dfrac{-5}{6} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = \left(-\dfrac{2}{3}\right) \times \left(\dfrac{3}{2}\right) = -1 = \dfrac{-6}{6} = \dfrac{c}{a} \quad \checkmark$
So, $\alpha = -\dfrac{2}{3}, \beta = \dfrac{3}{2}$.
Here, $a = 6, b = -5, c = -6$.
Verification:
Sum: $\alpha + \beta = -\dfrac{2}{3} + \dfrac{3}{2} = \dfrac{-4 + 9}{6} = \dfrac{5}{6} = -\dfrac{-5}{6} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = \left(-\dfrac{2}{3}\right) \times \left(\dfrac{3}{2}\right) = -1 = \dfrac{-6}{6} = \dfrac{c}{a} \quad \checkmark$
(iv) $4x^2 + 12x$
Factorise by taking common factor:
$$4x(x + 3) = 0 \implies x = 0 \text{ or } x = -3$$
So, $\alpha = 0, \beta = -3$.
Here, $a = 4, b = 12, c = 0$.
Verification:
Sum: $\alpha + \beta = 0 + (-3) = -3 = -\dfrac{12}{4} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = 0 \times (-3) = 0 = \dfrac{0}{4} = \dfrac{c}{a} \quad \checkmark$
Here, $a = 4, b = 12, c = 0$.
Verification:
Sum: $\alpha + \beta = 0 + (-3) = -3 = -\dfrac{12}{4} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = 0 \times (-3) = 0 = \dfrac{0}{4} = \dfrac{c}{a} \quad \checkmark$
(v) $t^2 - 20$
Factorise using difference of squares $a^2 - b^2 = (a - b)(a + b)$:
$$t^2 - 20 = (t - \sqrt{20})(t + \sqrt{20}) = (t - 2\sqrt{5})(t + 2\sqrt{5})$$
Setting $p(t) = 0 \implies t = 2\sqrt{5}$ or $t = -2\sqrt{5}$.
Here, $a = 1, b = 0, c = -20$.
Verification:
Sum: $\alpha + \beta = 2\sqrt{5} + (-2\sqrt{5}) = 0 = -\dfrac{0}{1} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = (2\sqrt{5})(-2\sqrt{5}) = -20 = \dfrac{-20}{1} = \dfrac{c}{a} \quad \checkmark$
Here, $a = 1, b = 0, c = -20$.
Verification:
Sum: $\alpha + \beta = 2\sqrt{5} + (-2\sqrt{5}) = 0 = -\dfrac{0}{1} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = (2\sqrt{5})(-2\sqrt{5}) = -20 = \dfrac{-20}{1} = \dfrac{c}{a} \quad \checkmark$
(vi) $3x^2 - 5x - 2$
Factorise: product = $-6$, sum = $-5 \implies -6$ and $1$:
$$3x^2 - 6x + x - 2 = 3x(x - 2) + 1(x - 2) = (3x + 1)(x - 2)$$
Setting $p(x) = 0 \implies x = -\dfrac{1}{3}$ or $x = 2$.
Here, $a = 3, b = -5, c = -2$.
Verification:
Sum: $\alpha + \beta = -\dfrac{1}{3} + 2 = \dfrac{5}{3} = -\dfrac{-5}{3} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = \left(-\dfrac{1}{3}\right) \times 2 = -\dfrac{2}{3} = \dfrac{c}{a} \quad \checkmark$
Here, $a = 3, b = -5, c = -2$.
Verification:
Sum: $\alpha + \beta = -\dfrac{1}{3} + 2 = \dfrac{5}{3} = -\dfrac{-5}{3} = -\dfrac{b}{a} \quad \checkmark$
Product: $\alpha\beta = \left(-\dfrac{1}{3}\right) \times 2 = -\dfrac{2}{3} = \dfrac{c}{a} \quad \checkmark$
Relationship Verified: $\alpha + \beta = -\dfrac{b}{a}$ and $\alpha\beta = \dfrac{c}{a}$ hold for all polynomials.
Question 2
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively:
(i) $\text{Sum} = \dfrac{1}{2}, \text{Product} = -2$
(ii) $\text{Sum} = \sqrt{3}, \text{Product} = \dfrac{1}{4}$
(iii) $\text{Sum} = 0, \text{Product} = 6$
(iv) $\text{Sum} = 2, \text{Product} = 1$ (v) $\text{Sum} = -\dfrac{1}{2}, \text{Product} = \dfrac{1}{3}$ (vi) $\text{Sum} = 5, \text{Product} = 6$
(iv) $\text{Sum} = 2, \text{Product} = 1$ (v) $\text{Sum} = -\dfrac{1}{2}, \text{Product} = \dfrac{1}{3}$ (vi) $\text{Sum} = 5, \text{Product} = 6$
Standard Formula
A quadratic polynomial with sum of zeroes $S$ and product of zeroes $P$ is given by:
$$p(x) = k\left[x^2 - Sx + P\right] \quad (k \neq 0)$$
(i) $S = \dfrac{1}{2}, P = -2$
$$p(x) = k\left[x^2 - \frac{1}{2}x - 2\right]$$
For $k = 2$, we obtain:
$$\mathbf{2x^2 - x - 4}$$
(ii) $S = \sqrt{3}, P = \dfrac{1}{4}$
$$p(x) = k\left[x^2 - \sqrt{3}x + \frac{1}{4}\right]$$
For $k = 4$, we obtain:
$$\mathbf{4x^2 - 4\sqrt{3}x + 1}$$
(iii) $S = 0, P = 6$
$$p(x) = k\left[x^2 - 0x + 6\right] = \mathbf{x^2 + 6}$$
(iv) $S = 2, P = 1$
$$p(x) = k\left[x^2 - 2x + 1\right] = \mathbf{x^2 - 2x + 1}$$
(v) $S = -\dfrac{1}{2}, P = \dfrac{1}{3}$
$$p(x) = k\left[x^2 - \left(-\frac{1}{2}\right)x + \frac{1}{3}\right] = k\left[x^2 + \frac{1}{2}x + \frac{1}{3}\right]$$
For $k = 6$, we obtain:
$$\mathbf{6x^2 + 3x + 2}$$
(vi) $S = 5, P = 6$
$$p(x) = k\left[x^2 - 5x + 6\right] = \mathbf{x^2 - 5x + 6}$$
Polynomials: (i) $2x^2 - x - 4$, (ii) $4x^2 - 4\sqrt{3}x + 1$, (iii) $x^2 + 6$, (iv) $x^2 - 2x + 1$, (v) $6x^2 + 3x + 2$, (vi) $x^2 - 5x + 6$.