Find the roots of the following quadratic equations by factorisation:
(i) \(x^2 - 3x - 10 = 0\)
(ii) \(2x^2 + x - 6 = 0\)
(iii) \(\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0\)
(iv) \(2x^2 - x + \dfrac{1}{8} = 0\)
(v) \(100x^2 - 20x + 1 = 0\)
(i) \(-2, 5\)
(ii) \(-2, \dfrac{3}{2}\)
(iii) \(-\dfrac{5}{\sqrt{2}}, -\sqrt{2}\)
(iv) \(\dfrac{1}{4}, \dfrac{1}{4}\)
(v) \(\dfrac{1}{10}, \dfrac{1}{10}\)
Method reminder: For each quadratic equation, we factorise the left-hand side and then use the zero product rule: if \((A)(B) = 0\), then \(A = 0\) or \(B = 0\).
1. We look for two numbers whose product is \(-10\) and whose sum is \(-3\). These numbers are \(-5\) and \(+2\) because \(-5 \times 2 = -10\) and \(-5 + 2 = -3\).
2. Split the middle term \(-3x\) using \(-5x\) and \(+2x\):
\(x^2 - 3x - 10 = x^2 - 5x + 2x - 10\).
3. Group the terms in pairs and factor each group:
\(x^2 - 5x + 2x - 10 = x(x - 5) + 2(x - 5)\).
4. Take \((x - 5)\) common:
\(x(x - 5) + 2(x - 5) = (x + 2)(x - 5)\).
5. So the equation becomes:
\((x + 2)(x - 5) = 0\).
6. Set each factor equal to zero:
\(x + 2 = 0 \Rightarrow x = -2\); \(x - 5 = 0 \Rightarrow x = 5\).
Roots: \(x = -2, 5\).
1. Here, the product of the first and last coefficients is \(2 \times (-6) = -12\). We need two numbers whose product is \(-12\) and sum is \(1\) (the coefficient of \(x\)). These numbers are \(4\) and \(-3\).
2. Split the middle term \(x\) as \(4x - 3x\):
\(2x^2 + x - 6 = 2x^2 + 4x - 3x - 6\).
3. Group the terms in pairs:
\(2x^2 + 4x - 3x - 6 = 2x(x + 2) - 3(x + 2)\).
4. Factor out \((x + 2)\):
\(2x(x + 2) - 3(x + 2) = (2x - 3)(x + 2)\).
5. Equation becomes:
\((2x - 3)(x + 2) = 0\).
6. Set each factor to zero:
\(2x - 3 = 0 \Rightarrow 2x = 3 \Rightarrow x = \dfrac{3}{2}\).
\(x + 2 = 0 \Rightarrow x = -2\).
Roots: \(x = -2, \dfrac{3}{2}\).
1. Here, the product of the first and last coefficients is \(\sqrt{2} \times 5\sqrt{2} = 5 \times 2 = 10\).
2. We need two numbers whose product is \(10\) and sum is \(7\). The numbers are \(5\) and \(2\).
3. We split \(7x\) as \(5\sqrt{2}x + \sqrt{2}x\), because their coefficients (\(5\sqrt{2}\) and \(\sqrt{2}\)) correspond to the chosen numbers and involve \(\sqrt{2}\):
\(\sqrt{2}x^2 + 7x + 5\sqrt{2} = \sqrt{2}x^2 + 5\sqrt{2}x + \sqrt{2}x + 5\sqrt{2}\).
4. Group and factorise:
First pair: \(\sqrt{2}x^2 + 5\sqrt{2}x = \sqrt{2}x(x + 5)\).
Second pair: \(\sqrt{2}x + 5\sqrt{2} = \sqrt{2}(x + 5)\).
So, \(\sqrt{2}x^2 + 5\sqrt{2}x + \sqrt{2}x + 5\sqrt{2} = (x + 5)(\sqrt{2}x + \sqrt{2})\).
5. Factor out \(\sqrt{2}\) from the second bracket:
\((x + 5)(\sqrt{2}x + \sqrt{2}) = (x + 5)\sqrt{2}(x + 1)\), but for finding roots it is simpler to keep it as:
\((\sqrt{2}x + 5)(x + \sqrt{2}) = 0\).
6. Set each factor equal to zero:
\(\sqrt{2}x + 5 = 0 \Rightarrow \sqrt{2}x = -5 \Rightarrow x = -\dfrac{5}{\sqrt{2}}\).
\(x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2}\).
Roots: \(x = -\dfrac{5}{\sqrt{2}}, -\sqrt{2}\).
1. To remove the fraction, multiply the whole equation by 8:
\(8(2x^2 - x + \dfrac{1}{8}) = 0 \Rightarrow 16x^2 - 8x + 1 = 0\).
2. Recognise this as a perfect square: \((4x)^2 = 16x^2\), \(-2 \cdot 4x \cdot 1 = -8x\), and \(1^2 = 1\). So:
\(16x^2 - 8x + 1 = (4x - 1)^2\).
3. Hence, we can write:
\((4x - 1)^2 = 0\).
4. Take square root on both sides (or directly set the factor equal to zero):
\(4x - 1 = 0 \Rightarrow 4x = 1 \Rightarrow x = \dfrac{1}{4}\).
5. Since the same factor appears twice, this is a repeated (double) root.
Root: \(x = \dfrac{1}{4}, \dfrac{1}{4}\).
1. Again, observe the pattern of a perfect square. Note that \((10x)^2 = 100x^2\), \(-2 \cdot 10x \cdot 1 = -20x\), and \(1^2 = 1\).
2. So, we can write:
\(100x^2 - 20x + 1 = (10x - 1)^2\).
3. Therefore, the equation becomes:
\((10x - 1)^2 = 0\).
4. Set the factor equal to zero:
\(10x - 1 = 0 \Rightarrow 10x = 1 \Rightarrow x = \dfrac{1}{10}\).
5. Here also, the factor is repeated, so there is a double root.
Root: \(x = \dfrac{1}{10}, \dfrac{1}{10}\).
Summary: By careful splitting of the middle term and recognising perfect-square patterns, we obtain all roots by factorisation.