Quadratic equations and graphs – Year 10 GCSE maths revision notes to download as PDF

Worked example 1

\(x^2-5x+6=0\).

Numbers with product 6 and sum \(-5\) are \(-2\) and \(-3\). \((x-2)(x-3)=0\), so \(x=2\) or \(x=3\).

Worked example 2

\(x^2=5x-4\).

Rearrange to \(x^2-5x+4=0\). \((x-1)(x-4)=0\), so \(x=1\) or \(x=4\). Never divide both sides by \(x\), because that loses a solution.

Worked example 3

Solve \(x^2+3x=10\) [3 marks].

Rearrange: \(x^2+3x-10=0\). Factorise \((x+5)(x-2)=0\). So \(x=-5\) or \(x=2\).

Worked example 4

The graph of \(y=x^2-4x+3\) crosses the \(x\)-axis at \(x=1\) and \(x=3\). Write down the equation of the line of symmetry and the coordinates of the minimum point [3 marks].

Line of symmetry \(x=2\) (midpoint of 1 and 3). At \(x=2\), \(y=4-8+3=-1\). Minimum point \((2,-1)\).

Worked example 5 (Higher tier)

Solve \(3x^2-5x-2=0\) by factorising [3 marks].

\(ac=-6\); numbers \(-6\) and \(1\). \(3x^2-6x+x-2=3x(x-2)+1(x-2)=(3x+1)(x-2)\). \(x=-\frac{1}{3}\) or \(x=2\).

Worked example 6 (Higher tier)

Use the quadratic formula to solve \(x^2-4x-3=0\) to 2 d.p. [3 marks].

\(x=\dfrac{4\pm\sqrt{16+12}}{2}=\dfrac{4\pm\sqrt{28}}{2}\), so \(x=4.65\) or \(x=-0.65\).

Worked example 7: finding the turning point

Find the turning point of \(y=x^2-8x+7\) using the roots [3 marks].

Factorise: \(y=(x-1)(x-7)\), so the roots are \(x=1\) and \(x=7\). The line of symmetry is halfway between them: \(x=\dfrac{1+7}{2}=4\). Substitute: \(y=16-32+7=-9\). The turning point is \((4,-9)\) and it is a minimum because the \(x^2\) coefficient is positive. You can also use completing the square: \(y=(x-4)^2-9\) gives the same point.

Worked example 8: reading a graph for an equation

The graph of \(y=x^2-4x-1\) is drawn. Explain how to use it to solve \(x^2-4x-1=3\) [2 marks].

Draw the horizontal line \(y=3\) on the same axes. The \(x\)-values of the points where the line crosses the curve are the solutions of \(x^2-4x-1=3\). This works because on the curve \(y=x^2-4x-1\) and on the line \(y=3\), so where they meet \(x^2-4x-1=3\). The solutions are \(x=2\pm\sqrt8\), about \(-0.8\) and \(4.8\).

Worked example 9: a quadratic with no real solutions

Show that \(x^2+2x+5=0\) has no real solutions [3 marks].

The discriminant is \(b^2-4ac=4-20=-16\), which is negative, so there are no real roots. By completing the square, \(x^2+2x+5=(x+1)^2+4\). A squared bracket is never negative, so the smallest value of the expression is 4, which means the graph stays above the \(x\)-axis and never crosses it.

Worked example 10: a problem in context

A rectangle is 3 cm longer than it is wide. A square of side 2 cm is removed from one corner, leaving an area of 36 cm². Find the width of the rectangle [4 marks].

Let the width be \(x\). The rectangle has area \(x(x+3)\). Remove 4 cm²: \(x(x+3)-4=36\), so \(x^2+3x-40=0\). Factorise: \((x+8)(x-5)=0\). The width is positive, so \(x=5\) cm. Check: the rectangle is \(5\times8=40\), less 4 gives 36. ✓