Edexcel Gcse Mathematics Linear 1ma0 Solving Quadratics By Factorising Answers

Solving quadratic equations

Solve quadratic equations by factorising, using formulae and completing the square. Each method also provides information about the corresponding quadratic graph.

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Quadratic equations

A quadratic equation contains terms up to \(x^2\) . There are many ways to solve quadratics. All quadratic equations can be written in the form \(ax^2 + bx + c = 0\) where \(a\) , \(b\) and \(c\) are numbers ( \(a\) cannot be equal to 0, but \(b\) and \(c\) can be).

Here are some examples of quadratic equations in this form:

\(2x^2 - 2x - 3 = 0\) . \(a = 2\) , \(b = -2\) and \(c = -3\)
\(2x(x + 3) = 0\) . \(a = 2\) , \(b = 6\) and \(c = 0\) (in this example, the bracket can be expanded to \(2x^2 + 6x = 0\) )
\((2x + 1)(x - 5) = 0\) . \(a = 2\) , \(b = -9\) and \(c = -5\) (this will expand to \(2x^2 - 9x - 5 = 0\) )
\(x^2 + 2 = 4\) . \(a = 1\) , \(b = 0\) and \(c = -2\)
\(3x^2 = 48\) . \(a = 3\) , \(b = 0\) and \(c = -48\) (in this example \(c = -48\) , but has been rearranged to the other side of the equation)

\(3x^2 = 48\) is an example of a quadratic equation that can be solved simply.

Square root both sides to isolate x

The positive and negative square roots of 16 are 4 and -4, as 4 × 4 = 16 and -4 × -4 = 16, so x = ± 4

Solving by factorising

If the product of two numbers is zero then one or both of the numbers must also be equal to zero. Therefore if \(ab = 0\) , then \(a = 0\) and/or \(b = 0\) .

If \((x + 1)(x + 2) = 0\) , then \(x + 1 = 0\) or \(x + 2 = 0\) , or both. Factorising quadratics will also be used to solve these equations.

Expanding the brackets \((x + 2)(x + 3)\) gives \(x^2 + 3x + 2x + 6\) , which simplifies to \(x^2 + 5x + 6\) . Factorising is the reverse process of expanding brackets, so factorising \(x^2 + 5x + 6\) gives \((x + 2)(x + 3)\) .

Example

Solve \(x(x + 3) = 0\) .

The product of \(x\) and \(x + 3\) is 0, so \(x = 0\) or \(x + 3 = 0\) , or both.

\[\begin{array}{rcl} x + 3 & = & 0 \\ -3 && -3 \\ x & = & -3 \end{array}\]

\(x = 0\) or \(x = -3\) .

Question

Solve \((x + 1)(x - 5) = 0\) .

The product of \(x + 1\) and \(x - 5\) is 0, so one or both brackets must also be equal to 0.

\[\begin{array}{rcl} x + 1 & = & 0 \\ -1 && -1 \\ x & = & -1 \end{array}\]

\[\begin{array}{rcl} x - 5 & = & 0 \\ + 5 && + 5 \\ x & = & 5 \end{array}\]

\(x = -1\) or \(x = 5\)

Question

Solve \(x^2 + 7x + 12 = 0\) .

Begin by factorising the quadratic .

The quadratic will be in the form \((x + a)(x + b) = 0\) .

Find two numbers with a product of 12 and a sum of 7.

\(3 \times 4 = 12\) , and \(3 + 4 = 7\) , so \(a\) and \(b\) are equal to 3 and 4. This gives:

\[(x + 3)(x + 4) = 0\]

The product of \(x + 3\) and \(x + 4\) is 0, so \(x + 3 = 0\) or \(x + 4 = 0\) , or both.

\[\begin{array}{rcl} x + 3 & = & 0 \\ - 3 && - 3 \\ x & = & -3 \end{array}\]

\[\begin{array}{rcl} x + 4 & = & 0 \\ - 4 && - 4 \\ x & = & -4 \end{array}\]

\(x = -3\) or \(x = -4\)

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