Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) \(x^2 - 2x - 8\)
(ii) \(4s^2 - 4s + 1\)
(iii) \(6x^2 - 3 - 7x\)
(iv) \(4u^2 + 8u\)
(v) \(t^2 - 15\)
(vi) \(3x^2 - x - 4\)
General relation between zeroes and coefficients
For a quadratic polynomial \(ax^2 + bx + c\) with zeroes \(\alpha\) and \(\beta\) (\(a \neq 0\)), the following relations hold:
\[ \alpha + \beta = -\dfrac{b}{a}, \qquad \alpha\beta = \dfrac{c}{a}. \]
These will be verified in each case.
(i) \(x^2 - 2x - 8\)
Consider the quadratic equation
\[ x^2 - 2x - 8 = 0. \]
Factorise the quadratic expression. We need two numbers whose product is \(-8\) and sum is \(-2\). These numbers are \(-4\) and \(2\).
\[ x^2 - 2x - 8 = x^2 - 4x + 2x - 8 = x(x - 4) + 2(x - 4) = (x - 4)(x + 2). \]
So,
\[ (x - 4)(x + 2) = 0 \Rightarrow x - 4 = 0 \text{ or } x + 2 = 0. \]
Hence the zeroes are
\[ x = 4, \; x = -2. \]
Let \(\alpha = 4\) and \(\beta = -2\).
Sum of zeroes: \(\alpha + \beta = 4 + (-2) = 2\).
Product of zeroes: \(\alpha\beta = 4 \times (-2) = -8\).
For \(x^2 - 2x - 8\), \(a = 1, b = -2, c = -8\).
\[ -\dfrac{b}{a} = -\dfrac{-2}{1} = 2, \quad \dfrac{c}{a} = \dfrac{-8}{1} = -8. \]
Thus, \(\alpha + \beta = -\dfrac{b}{a}\) and \(\alpha\beta = \dfrac{c}{a}\). Relation verified.
(ii) \(4s^2 - 4s + 1\)
Consider
\[ 4s^2 - 4s + 1 = 0. \]
Try to factorise:
\[ 4s^2 - 4s + 1 = (2s - 1)^2. \]
So
\[ (2s - 1)^2 = 0 \Rightarrow 2s - 1 = 0 \Rightarrow s = \dfrac{1}{2}. \]
Both zeroes are equal, \(\alpha = \beta = \dfrac{1}{2}\).
Sum of zeroes:
\[ \alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2} = 1. \]
Product of zeroes:
\[ \alpha\beta = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}. \]
Here \(a = 4, b = -4, c = 1\).
\[ -\dfrac{b}{a} = -\dfrac{-4}{4} = 1, \quad \dfrac{c}{a} = \dfrac{1}{4}. \]
Again, the relation is verified.
(iii) \(6x^2 - 3 - 7x\)
First write the polynomial in standard order of powers of \(x\):
\[ 6x^2 - 7x - 3. \]
Set it equal to zero:
\[ 6x^2 - 7x - 3 = 0. \]
Use the quadratic formula
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
Here \(a = 6, b = -7, c = -3\).
Discriminant:
\[ D = b^2 - 4ac = (-7)^2 - 4(6)(-3) = 49 + 72 = 121. \]
\(D = 121 = 11^2\).
Therefore,
\[ x = \dfrac{-(-7) \pm \sqrt{121}}{2 \cdot 6} = \dfrac{7 \pm 11}{12}. \]
So the two zeroes are
\[ x_1 = \dfrac{7 + 11}{12} = \dfrac{18}{12} = \dfrac{3}{2}, \quad x_2 = \dfrac{7 - 11}{12} = \dfrac{-4}{12} = -\dfrac{1}{3}. \]
Thus, \(\alpha = \dfrac{3}{2}\) and \(\beta = -\dfrac{1}{3}\).
Sum:
\[ \alpha + \beta = \dfrac{3}{2} + \left(-\dfrac{1}{3}\right) = \dfrac{9 - 2}{6} = \dfrac{7}{6}. \]
Product:
\[ \alpha\beta = \dfrac{3}{2} \times \left(-\dfrac{1}{3}\right) = -\dfrac{3}{6} = -\dfrac{1}{2}. \]
Now
\[ -\dfrac{b}{a} = -\dfrac{-7}{6} = \dfrac{7}{6}, \quad \dfrac{c}{a} = \dfrac{-3}{6} = -\dfrac{1}{2}. \]
Thus, the relation is satisfied.
(iv) \(4u^2 + 8u\)
Write the quadratic equation:
\[ 4u^2 + 8u = 0. \]
Factor out the common term:
\[ 4u^2 + 8u = 4u(u + 2). \]
So
\[ 4u(u + 2) = 0 \Rightarrow 4u = 0 \text{ or } u + 2 = 0. \]
Hence,
\[ u = 0, \quad u = -2. \]
Let \(\alpha = 0\) and \(\beta = -2\).
Sum:
\[ \alpha + \beta = 0 + (-2) = -2. \]
Product:
\[ \alpha\beta = 0 \times (-2) = 0. \]
Here, \(a = 4, b = 8, c = 0\).
\[ -\dfrac{b}{a} = -\dfrac{8}{4} = -2, \quad \dfrac{c}{a} = \dfrac{0}{4} = 0. \]
The relation holds.
(v) \(t^2 - 15\)
Consider
\[ t^2 - 15 = 0. \]
Move the constant term to the other side:
\[ t^2 = 15. \]
Taking square roots,
\[ t = \pm \sqrt{15}. \]
So the zeroes are \(\alpha = \sqrt{15}\) and \(\beta = -\sqrt{15}\).
Sum:
\[ \alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0. \]
Product:
\[ \alpha\beta = \sqrt{15} \times (-\sqrt{15}) = -15. \]
In \(t^2 - 15\), we have \(a = 1, b = 0, c = -15\).
\[ -\dfrac{b}{a} = -\dfrac{0}{1} = 0, \quad \dfrac{c}{a} = \dfrac{-15}{1} = -15. \]
Hence, the relation is verified.
(vi) \(3x^2 - x - 4\)
Set
\[ 3x^2 - x - 4 = 0. \]
Use the quadratic formula with \(a = 3, b = -1, c = -4\).
Discriminant:
\[ D = b^2 - 4ac = (-1)^2 - 4(3)(-4) = 1 + 48 = 49. \]
\(D = 49 = 7^2\).
So,
\[ x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-(-1) \pm 7}{2 \cdot 3} = \dfrac{1 \pm 7}{6}. \]
Thus, the two zeroes are
\[ x_1 = \dfrac{1 + 7}{6} = \dfrac{8}{6} = \dfrac{4}{3}, \quad x_2 = \dfrac{1 - 7}{6} = \dfrac{-6}{6} = -1. \]
Let \(\alpha = \dfrac{4}{3}\), \(\beta = -1\).
Sum:
\[ \alpha + \beta = \dfrac{4}{3} + (-1) = \dfrac{4 - 3}{3} = \dfrac{1}{3}. \]
Product:
\[ \alpha\beta = \dfrac{4}{3} \times (-1) = -\dfrac{4}{3}. \]
Here \(a = 3, b = -1, c = -4\).
\[ -\dfrac{b}{a} = -\dfrac{-1}{3} = \dfrac{1}{3}, \quad \dfrac{c}{a} = \dfrac{-4}{3}. \]
Thus, the relationship between zeroes and coefficients is verified in every case.