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Solving Quadratics by Factorising
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Foundational skills
Solve from factorised form (x + a)(x + b) = 0
\[ (x + 3)(x – 5) = 0 \]
Solve a quadratic already in factorised form.
Solve from factorised form (ax + b)(cx + d) = 0
\[ (2x + 3)(3x – 1) = 0 \]
Solve with non-unit x coefficients, giving fractions.
Solve x(x + a) = 0 where one solution is zero
\[ x(x – 4) = 0 \]
Solve where x is a common factor.
Monic quadratics in standard form
Solve x² + bx + c = 0 with two positive solutions
\[ x^2 – 7x + 12 = 0 \]
Both solutions are positive.
Solve x² + bx + c = 0 with two negative solutions
\[ x^2 + 7x + 12 = 0 \]
Both solutions are negative.
Solve x² + bx + c = 0 with solutions of opposite signs
\[ x^2 + 2x – 15 = 0 \]
One positive and one negative solution.
Solve x² + bx = 0 where c = 0
\[ x^2 + 5x = 0 \]
Factorise out x. Do not divide by x!
Monic quadratics requiring rearrangement
Solve x² + bx = c by rearranging
\[ x^2 – 5x = 6 \]
Constant is on the wrong side.
Solve x² = bx + c by rearranging
\[ x^2 = 3x + 10 \]
All terms except x² on wrong side.
Solve quadratics with terms on both sides
\[ x^2 + 3x – 5 = 2x + 7 \]
Collect terms from both sides first.
Expand x(x + a) = b then solve
\[ x(x + 3) = 10 \]
Expand, rearrange, then factorise.
Expand (x + a)(x + b) = c then solve
\[ (x + 1)(x + 2) = 12 \]
Expand two brackets first.
Expand (x + a)² = b then solve
\[ (x + 3)^2 = 16 \]
Expand squared bracket first.
Non-monic quadratics in standard form
Solve ax² + bx + c = 0 where a = 2
\[ 2x^2 + 5x + 3 = 0 \]
Coefficient of x² is 2.
Solve ax² + bx + c = 0 where a > 2
\[ 3x^2 + 11x + 6 = 0 \]
Coefficient of x² is 3 or more.
Solve ax² + bx = 0 by extracting common factor
\[ 3x^2 + 12x = 0 \]
No constant term. Factor out x.
Solve -x² + bx + c = 0 (negative leading coefficient)
\[ -x^2 + 5x + 6 = 0 \]
The x² term is negative.
Solve ax² + bx + c = 0 with negative a < -1
\[ -2x^2 + 7x + 4 = 0 \]
Negative coefficient other than -1.
Non-monic quadratics requiring rearrangement
Solve ax² + bx = c by rearranging
\[ 2x^2 + 5x = 3 \]
Non-monic with constant on wrong side.
Solve ax² + c = bx by rearranging
\[ 3x^2 + 2 = 7x \]
Non-monic with x term on wrong side.
Expand x(ax + b) = c then solve
\[ x(2x + 3) = 5 \]
Expand bracket with coefficient first.
Expand (ax + b)(cx + d) = e then solve
\[ (2x + 1)(x + 3) = 4 \]
Expand non-monic brackets first.
Difference of two squares
Solve x² – a² = 0
\[ x^2 – 25 = 0 \]
Difference of two squares pattern.
Solve x² = a²
\[ x^2 = 49 \]
x² equals a perfect square.
Solve ax² – b² = 0 (both perfect squares)
\[ 4x^2 – 9 = 0 \]
Coefficient of x² is also a perfect square.
Solve ax² = b where b/a is a perfect square
\[ 3x^2 = 27 \]
Divide first, then square root.
Common factor to extract
Solve by extracting a common factor first
\[ 2x^2 + 8x + 6 = 0 \]
All terms share a common factor.
Solve ax² + bx = 0 extracting full common factor
\[ 6x^2 + 15x = 0 \]
Extract HCF including numerical factor.
Perfect squares (repeated roots)
Solve x² + 2ax + a² = 0 (perfect square trinomial)
\[ x^2 + 6x + 9 = 0 \]
Perfect square gives one repeated root.
Solve non-monic perfect square
\[ 4x^2 + 12x + 9 = 0 \]
Non-monic perfect square trinomial.
Hidden quadratics
Solve x⁴ + bx² + c = 0 by substitution
\[ x^4 – 5x^2 + 4 = 0 \]
Substitute u = x² to create a quadratic.
Solve x⁴ + bx² + c = 0 (one u-value rejected)
\[ x^4 + 3x^2 – 4 = 0 \]
One substitution result is negative.
Special cases
Solve with fractional coefficients
\[ \frac{1}{2}x^2 + x – \frac{3}{2} = 0 \]
Clear fractions first by multiplying.
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