Find the roots of the following quadratic equations by the factorisation method:
(i) \(2x^2 + \dfrac{5}{3}x - 2 = 0\)
(ii) \(\dfrac{2}{5}x^2 - x - \dfrac{3}{5} = 0\)
(iii) \(3\sqrt{2}\,x^2 - 5x - \sqrt{2} = 0\)
(iv) \(3x^2 + 5\sqrt{5}\,x - 10 = 0\)
(v) \(21x^2 - 2x + \dfrac{1}{21} = 0\)
(i) \(x=-\dfrac{3}{2}\), \(x=\dfrac{2}{3}\)
(ii) \(x=-\dfrac{1}{2}\), \(x=3\)
(iii) \(x=\sqrt{2}\), \(x=-\dfrac{\sqrt{2}}{6}\)
(iv) \(x=\dfrac{\sqrt{5}}{3}\), \(x=-2\sqrt{5}\)
(v) double root \(x=\dfrac{1}{21}\)
Equation: \(2x^2 + \dfrac{5}{3}x - 2 = 0\)
Step 1: Remove the fraction by multiplying everything by 3:
\(6x^2 + 5x - 6 = 0\)
Step 2: Multiply first and last coefficient: \(6 \times -6 = -36\).
We need two numbers whose product is -36 and sum is +5. These are +9 and -4.
Step 3: Split middle term:
\(6x^2 + 9x - 4x - 6 = 0\)
Step 4: Group terms:
\((6x^2 + 9x) - (4x + 6) = 0\)
Step 5: Factor each group:
\(3x(2x + 3) - 2(2x + 3) = 0\)
Step 6: Take common factor:
\((2x + 3)(3x - 2) = 0\)
So, \(x = -\tfrac{3}{2}\) or \(x = \tfrac{2}{3}\).
Equation: \(\dfrac{2}{5}x^2 - x - \dfrac{3}{5} = 0\)
Step 1: Remove fraction by multiplying by 5:
\(2x^2 - 5x - 3 = 0\)
Step 2: Multiply first and last coefficient: \(2 \times -3 = -6\).
We need two numbers with product -6 and sum -5. These are -6 and +1.
Step 3: Split middle term:
\(2x^2 - 6x + x - 3 = 0\)
Step 4: Group terms:
\((2x^2 - 6x) + (x - 3) = 0\)
Step 5: Factor:
\(2x(x - 3) + 1(x - 3) = 0\)
Step 6: Take common factor:
\((2x + 1)(x - 3) = 0\)
So, \(x = -\tfrac{1}{2}\) or \(x = 3\).
Equation: \(3\sqrt{2}x^2 - 5x - \sqrt{2} = 0\)
Step 1: Multiply first and last coefficient: \(3\sqrt{2} \times -\sqrt{2} = -6\).
We need two numbers with product -6 and sum -5. These are -6 and +1.
Step 2: Split middle term:
\(3\sqrt{2}x^2 - 6x + x - \sqrt{2} = 0\)
Step 3: Group terms:
\((3\sqrt{2}x^2 - 6x) + (x - \sqrt{2}) = 0\)
Step 4: Factor:
\(3x(\sqrt{2}x - 2) + 1(x - \sqrt{2}) = 0\)
Step 5: Rearrange grouping:
\((3\sqrt{2}x + 1)(x - \sqrt{2}) = 0\)
So, \(x = \sqrt{2}\) or \(x = -\tfrac{\sqrt{2}}{6}\).
Equation: \(3x^2 + 5\sqrt{5}x - 10 = 0\)
Step 1: Look for possible factors with surds (square roots).
Try: \((3x - \sqrt{5})(x + 2\sqrt{5})\).
Step 2: Expand to check:
\(3x^2 + 6x\sqrt{5} - x\sqrt{5} - 10\)
= \(3x^2 + 5\sqrt{5}x - 10\). Correct!
Step 3: So equation is \((3x - \sqrt{5})(x + 2\sqrt{5}) = 0\).
Hence, \(x = \tfrac{\sqrt{5}}{3}\) or \(x = -2\sqrt{5}\).
Equation: \(21x^2 - 2x + \dfrac{1}{21} = 0\)
Step 1: Multiply everything by 21 to remove fraction:
\(441x^2 - 42x + 1 = 0\)
Step 2: Notice this looks like a perfect square.
\((21x - 1)^2 = 441x^2 - 42x + 1\)
Step 3: So equation is \((21x - 1)^2 = 0\).
That means root is \(x = \tfrac{1}{21}\), repeated twice (double root).