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Summary

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.

$\boxed{b^2-4ac}$

Inequality/Equation	Definition
$b^2 -4ac < 0$	No real solutions
$b^2 - 4ac = 0$	One real solution
$b^2-4ac > 0$	Two real solutions

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 $y = mx+2$ .

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