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:
\[ \alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}. \]
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 \(x^2 - 2x - 8\), \(a = 1, b = -2, c = -8\).
\[ -\frac{b}{a} = -\frac{-2}{1} = 2, \quad \frac{c}{a} = \frac{-8}{1} = -8. \]
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
\[ (2s - 1)^2 = 0 \Rightarrow 2s - 1 = 0 \Rightarrow s = \frac{1}{2}. \]
Both zeroes are equal, \(\alpha = \beta = \dfrac{1}{2}\).
Sum of zeroes:
\[ \alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1. \]
Product of zeroes:
\[ \alpha\beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}. \]
Here \(a = 4, b = -4, c = 1\).
\[ -\frac{b}{a} = -\frac{-4}{4} = 1, \quad \frac{c}{a} = \frac{1}{4}. \]
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
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \]
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,
\[ x = \frac{-(-7) \pm \sqrt{121}}{2 \cdot 6} = \frac{7 \pm 11}{12}. \]
So the two zeroes are
\[ x_1 = \frac{7 + 11}{12} = \frac{18}{12} = \frac{3}{2}, \quad x_2 = \frac{7 - 11}{12} = \frac{-4}{12} = -\frac{1}{3}. \]
Thus, \(\alpha = \dfrac{3}{2}\) and \(\beta = -\dfrac{1}{3}\).
Sum:
\[ \alpha + \beta = \frac{3}{2} + \left(-\frac{1}{3}\right) = \frac{9 - 2}{6} = \frac{7}{6}. \]
Product:
\[ \alpha\beta = \frac{3}{2} \times \left(-\frac{1}{3}\right) = -\frac{3}{6} = -\frac{1}{2}. \]
Now
\[ -\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}, \quad \frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}. \]
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, \(a = 4, b = 8, c = 0\).
\[ -\frac{b}{a} = -\frac{8}{4} = -2, \quad \frac{c}{a} = \frac{0}{4} = 0. \]
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 \(a = 1, b = 0, c = -15\).
\[ -\frac{b}{a} = -\frac{0}{1} = 0, \quad \frac{c}{a} = \frac{-15}{1} = -15. \]
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,
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-1) \pm 7}{2 \cdot 3} = \frac{1 \pm 7}{6}. \]
Thus, the two zeroes are
\[ x_1 = \frac{1 + 7}{6} = \frac{8}{6} = \frac{4}{3}, \quad x_2 = \frac{1 - 7}{6} = \frac{-6}{6} = -1. \]
Let \(\alpha = \dfrac{4}{3}\), \(\beta = -1\).
Sum:
\[ \alpha + \beta = \frac{4}{3} + (-1) = \frac{4 - 3}{3} = \frac{1}{3}. \]
Product:
\[ \alpha\beta = \frac{4}{3} \times (-1) = -\frac{4}{3}. \]
Here \(a = 3, b = -1, c = -4\).
\[ -\frac{b}{a} = -\frac{-1}{3} = \frac{1}{3}, \quad \frac{c}{a} = \frac{-4}{3}. \]
Thus, the relationship between zeroes and coefficients is verified in every case.
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) \(\dfrac{1}{4}, -1\)
(ii) \(\sqrt{2}, \dfrac{1}{3}\)
(iii) \(0, \sqrt{5}\)
(iv) \(1, 1\)
(v) \(-\dfrac{1}{4}, \dfrac{1}{4}\)
(vi) \(4, 1\)
Method using sum and product of zeroes
Let a quadratic polynomial have zeroes \(\alpha\) and \(\beta\). Then
\[ \alpha + \beta = S, \quad \alpha\beta = P. \]
The standard quadratic polynomial whose zeroes are \(\alpha\) and \(\beta\) is
\[ p(x) = x^2 - Sx + P. \]
Any non-zero scalar multiple of this polynomial represents the same pair of zeroes. The given first number is taken as the sum \(S\) and the second as the product \(P\).
(i) \(S = \dfrac{1}{4}, \; P = -1\)
Start with
\[ p(x) = x^2 - \frac{1}{4}x - 1. \]
To eliminate fractions, multiply the entire polynomial by 4 (a non-zero constant):
\[ 4p(x) = 4x^2 - x - 4. \]
So a convenient quadratic polynomial is
\[ p(x) = 4x^2 - x - 4. \]
This has sum of zeroes \(\dfrac{1}{4}\) and product \(-1\).
(ii) \(S = \sqrt{2}, \; P = \dfrac{1}{3}\)
Take
\[ p(x) = x^2 - \sqrt{2}x + \frac{1}{3}. \]
Multiply through by 3 to clear denominator:
\[ 3p(x) = 3x^2 - 3\sqrt{2}x + 1. \]
Thus one suitable polynomial is
\[ p(x) = 3x^2 - 3\sqrt{2}x + 1. \]
Its sum of zeroes is \(\sqrt{2}\) and product is \(\dfrac{1}{3}\).
(iii) \(S = 0, \; P = \sqrt{5}\)
Using the general form,
\[ p(x) = x^2 - Sx + P = x^2 - 0 \cdot x + \sqrt{5} = x^2 + \sqrt{5}. \]
Here there is no need to multiply by any constant, so
\[ p(x) = x^2 + \sqrt{5} \]
is the required quadratic polynomial whose zeroes have sum 0 and product \(\sqrt{5}\).
(iv) \(S = 1, \; P = 1\)
The polynomial is
\[ p(x) = x^2 - Sx + P = x^2 - x + 1. \]
The sum of its zeroes is 1 and product is 1, as required.
(v) \(S = -\dfrac{1}{4}, \; P = \dfrac{1}{4}\)
First write
\[ p(x) = x^2 - Sx + P = x^2 - \left(-\frac{1}{4}\right)x + \frac{1}{4} = x^2 + \frac{1}{4}x + \frac{1}{4}. \]
Multiply by 4 to remove fractions:
\[ 4p(x) = 4x^2 + x + 1. \]
So one convenient polynomial is
\[ p(x) = 4x^2 + x + 1. \]
This polynomial has sum of zeroes \(-\dfrac{1}{4}\) and product \(\dfrac{1}{4}\).
(vi) \(S = 4, \; P = 1\)
From the formula,
\[ p(x) = x^2 - Sx + P = x^2 - 4x + 1. \]
So
\[ p(x) = x^2 - 4x + 1 \]
is a quadratic polynomial whose zeroes have sum 4 and product 1.
Therefore, suitable quadratic polynomials for each pair \((S, P)\) are:
(i) \(4x^2 - x - 4\), (ii) \(3x^2 - 3\sqrt{2}x + 1\), (iii) \(x^2 + \sqrt{5}\), (iv) \(x^2 - x + 1\), (v) \(4x^2 + x + 1\), (vi) \(x^2 - 4x + 1\).