Using the Discriminant to Solve Simultaneous Equations – A Level Maths Revision

At A Level, you need to be able to solve simultaneous equations where one equation is non-linear. While a pair of linear equations always have a pair of solutions, for non-linear equations this is not necessarily the case. Using the discriminant will help you work out how many solutions there are – the following blog will explain how to do this.

The content below is also available in presentable and printable formats. If you would like to practice some of the prior-knowledge required for this topic, a multiple-choice quiz (with answers) is also available.

The solutions to a simultaneous equation are where the two equations would intersect if plotted on a graph. When dealing with a pair of linear equations, there is only ever one intersection. However, when one equation is linear and the other is quadratic there may be two, one or no points of intersection. This can be demonstrated by sketching the three possibilities:

You can use the discriminant, b2 – 4ac, to decide how many solutions there are to a pair of simultaneous equations.

Note that, if a straight line intersects a quadratic graph or a circle once, that line is a tangent to the curve at the point of intersection.

Example Question 1

Show that the simultaneous equations below have no real solution:
\begin{aligned} &y = x^2 - x + 2 \\ &2x + 3y = 4 \end{aligned}

Start by using substitution to get an equation with only x or y. In this case, the first equation is in the form y = …, so it will be simpler to substitute this into the second equation:

\begin{aligned} &2x + 3(x^2 - x + 2) = 4 \\ &2x + 3x^2 - 3x + 6 = 4 \\ &3x^2 - x + 2 = 0 \end{aligned}

You now have a quadratic equation with a single variable, in the form ax2 + bx + c = 0:

\begin{aligned} b^2 - 4ac &= (-1)^2 - 4 \times 3 \times 2 \\ &= 1 - 24 \\ &= -23 \end{aligned}

-23 is less than 0, so 3x2x + 2 = 0 has no real solutions.

Example Question 2

Find the number of points of intersection of:
\begin{aligned} &x^2 + y^2 = 36 \\ &3x + 2y = 8 \end{aligned}

This is slightly different to the previous example, as the first equation is a circle. However, it is still possible to have 0, 1 or 2 solutions, as seen below:

As before, start by rearranging and substituting the equations to find an equation with a single variable.

\begin{aligned} &3x + 2y = 8 \\ &3x = 8 - 2y \\ &x = \frac{8}{3} - \frac{2y}{3} \end{aligned}

You can now substitute into the first equation:

\begin{aligned} &(\frac{8}{3} -\frac{2y}{3})^2 + y^2 = 36 \\ &\frac{64}{9}-\frac{32y}{9}+\frac{4y^2}{9}+y^2 = 36 \\ & \frac{13y^2}{9} - \frac{32y}{9} - \frac{260}{9} = 0 \\ &13y^2 - 32y - 260 = 0 \end{aligned}

Once the equation is in the form ay2 + by + c = 0, you can find the discriminant:

\begin{aligned} b^2 - 4ac &= (-32)^2 - 4 \times 13 \times -260 \\ &= 14544 \end{aligned}

14544 > 0, therefore there are two solutions.

Practice Questions

1. Find the number of points of intersection for y = 4x2 – 5x + 3 and 2x – 3y = 7.

2. Find the number of points of intersection for 2(x – 3)2+3y2 = 12 and y = 8 – 5x.

3. Show that the line y = 7x – 55 is a tangent to the circle (x – 1)2 + (y – 2)2 = 50.

4. Find the values of c such that the graph of y = 2x + c is a tangent to y2 + x2 = 15

5. Consider the graphs of y = x2 – 3x – 3 and y = mx – 4. Find the values of m such that the graphs intercept at 2 points.

1.

\begin{aligned} &2x - 3y = 7 \therefore y = \frac{2x-7}{3} \\ &\frac{2x-7}{3}=4x^2-5x+3 \\ &2x-7=12x^2-15x+9 \\ &12x^2 - 17x + 16 = 0 \end{aligned}

\begin{aligned} b^2-4ac &= (-17)^2 - 4 \times 12 \times 16 \\ &= -479 \end{aligned}

-479 < 0 so there are no points of intersection.

2.

\begin{aligned} &2(x-3)^2 + 3(8-5x)^2 = 12 \\ &2x^2 - 12x + 18 + 75x^2 - 240x + 196 = 12 \\ &77x^2 - 252x + 202 = 0 \end{aligned}

\begin{aligned} b^2 - 4ac &= (-252)^2 - 4 \times 77 \times 202 \\ &= 1288 \end{aligned}

1288 > 0 so there are two points of intersection.

3.

\begin{aligned} &(x-1)^2 + (7x - 57)^2 = 50 \\ &x^2 -2x + 1 + 49x^2 - 798x + 3249 = 50 \\ &50x^2 - 800x + 3200 = 0 \end{aligned}

\begin{aligned} b^2 - 4ac &= (-800)^2 - 4\times 50 \times 3200 \\ &=0 \end{aligned}

Therefore, there is exactly one solution and y = 7x – 55 is a tangent to the circle.

4.

\begin{aligned} &(2x + c)^2 + x^2 = 15 \\& 4x^2 + 4cx + c^2 + x^2 = 15 \\ &5x^2 + 4cx + (c^2 - 15) = 0 \end{aligned}

\begin{aligned} b^2 - 4ac &= (4c)^2 - 4 \times 5 \times (c^2 - 15)\\ &= 16c^2 - 20c^2 + 300 \\&= -4c^2 + 300 \end{aligned}

\begin{aligned} &b^2-4ac =0 \\ & -4c^2 + 300 = 0 \\ &4c^2 = 300 \\ &c^2 = 75 \\ &c=\pm 5 \sqrt{3} \end{aligned}

5.

\begin{aligned} &x^2 - 3x - 3 = mx - 4 \\ &x^2 - (m+3)x + 1 = 0 \end{aligned}

\begin{aligned} b^2 - 4ac &= (-(m + 3))^2 - 4 \times 1 \times 1 \\ &= m^2 + 6m + 9 - 4 \\ &=m^2 + 6m + 5 \\ &=(m+1)(m+5) \end{aligned}

\begin{aligned} &b^2 - 4ac > 0 \\&(m+1)(m+5)>0 \\&m<-5 \text{ or } m>-1 \end{aligned}

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