Using the discriminant

​​In a nutshell

The discriminant identifies the number of solutions of a quadratic equation in the form $$ax^2 +bx +c = 0$$ where $$a \ne 0$$. If a quadratic equation can be formed from graphs' equations, the number of intersections can be found by comparing the value of the discriminant to $$0$$​. Take note that the solutions of graphs' simultaneous equations are the coordinates of their points of intersection on a graph. ​

The discriminant 

The discriminant is part of the quadratic formula: $$b^2 -4ac.$$ It can be used to find the number of solutions to a quadratic equation.

​

$$\boxed{b^2-4ac}$$​​

The discriminant from simultaneous equations

The solution of a simultaneous equation is the same as their points of intersection. By using substitution to form a quadratic equation then finding the discriminant, the number of points of intersection can be found. 

​​​​Example 1

How many points of intersections are there between the line $$y=2x-4$$ and the curve $$y = (x-2)^2$$?

​$$\begin{aligned} y&=2x+1 \quad\space\space\,(1)\\ y &= (x-2)^2 \quad (2) \end{aligned}$$​​

Substitute  $$y=2x +1$$ into equation 2.

​$$2x+1 = (x-2)^2$$​​

Collect the $$x$$ terms to one side. ​

​$$\begin{aligned} 2x+1 &= x^2 - 4x +4 \\ 0 &= x^2 -6x +3 \end{aligned}$$​​

Identify the constants $$a, b$$ and $$c$$.

$$\begin{aligned} 0&= ax^2 + bx +c \\ 0 &= 1x^2 -6x +3 \end{aligned}$$​​

​$$\begin{aligned} a&= 1 \\ b&=-6 \\ c&= 3 \end{aligned}$$

​

Substitute the values into the discriminant.

​$$\begin{aligned} b^2 &- 4ac \\ (-6)^2 -4 (1)(3) &= 36 - 12 = 24 \end{aligned}$$​​

Identify the number of intersections. 

$$b^2 - 4ac = 24 > 0$$

​​

Therefore, there are two points of intersection between $$y=2x-4$$ and $$y=(x-2)^2$$.

Example 2

There is one point of intersection between $$4x - 2y = 2$$ and $$x^2 +4ky + 5k = 0$$. What are the possible values of $$k$$?

​$$\begin{aligned} 4x-2y&=2 \quad\space\space(1)\\ x^2+4ky+5k &= 0 \space\space\quad (2) \end{aligned}$$​​

Rearrange equation $$1$$ to find $$y$$ in terms of $$x$$​. ​​

​$$\begin{aligned} 4x-2y&=2 \\ 4x-2 &=2y \\ 2x -1&=y\end{aligned}$$​​

​

Substitute $$2x-1 = y$$ into equation $$2$$.​​

​$$\begin{aligned}x^2 + 4k(2x-1) +5k &= 0 \\ x^2 + 8kx -4k + 5k &=0 \\ x^2 +8kx +k &=0 \end{aligned}$$​​

Identify the constants $$a,b$$ and $$c$$.

​$$\begin{aligned} 0&= ax^2 + bx +c \\ 0 &= 1x^2 +8kx +k \end{aligned}$$

$$\begin{aligned} a &= 1 \\b &= 8k \\c &= k \end{aligned}$$​​

If there is one point of intersection the discriminant is equal to $$0$$. 

Substitute the values into the discriminant.

​$$\begin{aligned} b^2 - 4ac &=0 \\ (8k)^2 -4 (1)(k) &= 64k^2 - 4k \\ 64k^2 - 4k &=0 \\ 4k(16k - 1)&=0 \end{aligned}$$​​

​

Solve for $$k$$.

​$$\begin{aligned} 4k &= 0 \\ k &= 0 \\ \\ 16k-1 &=\,\,0 \\ 16k & = \,\,1 \\ k &= \dfrac {1}{16} \end{aligned}$$​

 $$\underline{ k=0 \ or \ k=\dfrac{1}{16}}$$

Straight lines and circles

Circles have the equation: $$(x-a)^2 + (y-b)^2 = r^2$$, where the centre is $$(a,b)$$​ and radius $$r$$. The discriminant can be used to find whether a straight line intersects a circle.

​$$\boxed{(x-a)^2 + (y-b)^2=r^2}$$​​

Example 3

The line $$y = mx + 2$$​ is a tangent to the curve with equation $$x^2-5x+y^2 - 2y+4 = 0$$. What are the possible values of $$m$$?​

A tangent will have $$1$$ point of intersection with a circle. This means:

$$b^2-4ac = 0$$​​

Substitute the equation of the line into the equation of the circle to form a quadratic equation in terms of $$x$$:

​$$x^2 - 5x + (mx+2)^2 - 2(mx+2) + 4 = 0$$

​​

Simplify the quadratic equation by collecting terms:

​$$\begin{aligned}x^2 - 5x + (m^2x^2+4mx+4) - 2mx-4 + 4 &= 0 \\x^2 +m^2x^2- 5x +4mx- 2mx-4 +4+ 4 &= 0 \\ x^2 +m^2x^2- 5x +2mx+ 4 &= 0 \\ (1+m^2)x^2 +(2m-5)x+4 &= 0\end{aligned}$$​​

​​​​​​​​

Identify $$a$$, $$b$$ and $$c$$:

​$$\begin{aligned} a&=1+m^2 \\ b &= 2m-5 \\ c &= 4 \end{aligned}$$​​

Substitute the values into the discriminant:

​$$(2m-5)^2 - 4(1+m^2)(4) = 0$$​​

​

Simplify the equation:

$$(4m^2-20m+25)-(16+16m^2) = 0$$​​

​$$\begin{aligned}-12m^2 - 20m + 9 = 0 \\ 12m^2+20m - 9 = 0 \end{aligned}$$​​

​​

Solve for $$m$$ using one of the methods to solve quadratics such as completing the square:​

$$\begin{aligned}12m^2 + 20m - 9 &= 0 \\ \\m^2+\dfrac{20}{12}m - \dfrac9{12} &= 0 \\ \\ \left(m+\dfrac{10}{12}\right)^2 - \dfrac{100}{144} - \dfrac{9}{12} &= 0 \\ \\ \left(m+ \dfrac5{6}\right)^2 &= \dfrac{13}9 \\ \\ m + \dfrac56 &= \pm\sqrt{\dfrac{13}9} \\ \\ m &= \pm\dfrac{\sqrt{13}}{3} - \dfrac56 \end{aligned}$$​​

Therefore, $$\underline{m = \dfrac{\sqrt{13}}{3} - \dfrac56}$$ or $$\underline{ m = -\dfrac{\sqrt{13}}{3} - \dfrac56 }$$.​

Note: In this question, there are two possible tangents in the form $$y = mx+2$$.