Quadratics - Example

Question: Solve the quadratic equation \(x^2 - 5x + 6 = 0\).

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\(x^2 - 5x + 6 = 0\)	Write the quadratic equation in standard form.
\(-2 \times -3 = +6\), \(-2 + -3 = -5\)	Find two numbers that multiply to give \(+6\) and add to give \(-5\).
\((x - 2)(x - 3) = 0\)	Factorise the quadratic.
\(x - 2 = 0 \quad\) or \(\quad x - 3 = 0\)	Set each factor equal to zero.
\(x = 2\) or \(x = 3\)	Final Answer.

Question: Solve \(6x^2 + 5 = 17x\)

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\(6x^2 + 5 = 17x\)	Write in the form \(ax^2 + bx + c = 0\).
\(6x^2 - 17x + 5 = 0\)	Factorise.
\((2x - 5)(3x - 1) = 0\)	Use the fact that if \(pq = 0\), then \(p = 0\) or \(q = 0\).
\(2x - 5 = 0\) or \(3x - 1 = 0\)	Solve.
\(x = \tfrac{5}{2}\) or \(x = \tfrac{1}{3}\)	Final Answer.

Question: Solve \(9x^2 - 39x - 30 = 0\)

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\(9x^2 - 39x - 30 = 0\)	Divide both sides by the common factor of 3.
\(3x^2 - 13x - 10 = 0\)	Factorise.
\((3x + 2)(x - 5) = 0\)	Use the fact that if \(pq = 0\), then \(p = 0\) or \(q = 0\).
\(3x + 2 = 0\) or \(x - 5 = 0\)	Solve.
\(x = -\tfrac{2}{3}\) or \(x = 5\)	Final Answer.

Question: Solve \(\dfrac{21}{2x} - \dfrac{2}{x+3} = 1\).

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\(\dfrac{21}{2x} - \dfrac{2}{x+3} = 1\)	Multiply both sides by \(2x(x+3)\).
\(21(x+3) - 4x = 2x(x+3)\)	Expand brackets and rearrange.
\(2x^2 - 11x - 63 = 0\)	Simplify to standard quadratic form.
\((2x + 7)(x - 9) = 0\)	Factorise.
\(2x + 7 = 0\) or \(x - 9 = 0\)	Solve each factor.
\(x = -\tfrac{7}{2}\) or \(x = 9\)	Final Answer.
💡 Remember to check that your solutions do not make a denominator zero. Here, \(x \neq 0\) and \(x \neq -3\), so both answers are valid.

Question: Solve \(\dfrac{3x^2 + 26x + 35}{x^2 + 8} = 0\).

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Solving quadratic equations by factorisation