The Discriminant - 3 Rules For Success
The Discriminant – Introduction
At GCSE you will have come across the quadratic formula which you will know as:
x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}The part which is underneath the square root is referred to as the discriminant and using this helps us to deal with different types of questions involving quadratics as it tells you about the nature of such quadratics.
Using the discriminant to sketch graphs
There are three very distinct rules when it comes to the discriminant and it is important that you know them.
If b^2-4 a c>0
If this is the case then the sketch of a quadratic will look as follows:
You can see that the curve crosses the x-axis in two distinct places and so there are two distinct roots.
If b^2-4 a c=0
If this is the case then the sketch of a quadratic will look as follows:
You can see that the curve bounces off the x-axis as one specific value, this is known as a repeated root.
If b^2-4 a c<0
If this is the case then a sketch of a quadratic will look as follows:
You can see that the curve does not cross the x-axis at all and here the quadratic has no solutions.
The Discriminant - Examples
What is important is to be able to translate the English text into mathematical understanding. You will notice that in no cases does the question tell you to use the discriminant, but you need to.
The keywords that you need to translate are “has equal roots”. You will know that this means b^2-4 a c=0
For this example, a=1, b=k \text { and } c=9
And it is case of substituting these values into the discriminant as follows:
\begin{aligned} & k^2-4 \times 1 \times 9=0 \\ & k^2-36=0 \\ & k^2=36 \\ & \text { so } k= \pm 6 \end{aligned}
Again for this question it is important to be able to spot the key words within the question which are “distinct real solutions” and this means b^2-4 a c>0
For this particular example a=1, b=4 \text { and } c=k \text {. }
And so substituting these values into the discriminant you then have a inequality that needs to be solved as follows:
\begin{aligned} & 4^2-4 \times 1 \times k>0 \\ & 16-4 k>0 \\ & 16>4 k \\ & 4>k \\ & \text { So } k<4 \end{aligned}
The Discriminant – Exam Style Questions
The keywords here are “no real roots” and this means that b^2-4 a c<0
For this particular question a=1 \quad b=k \quad c=2
Substituting these values into the discriminant and solving the inequality gives the following:
\begin{aligned} & k^2-4(1)(2)<0 \\ & k^2-8<0<0 \\ & k^2<8 \\ & k=\sqrt{8} \quad k=-\sqrt{8} \end{aligned}It is always a good idea to sketch the curve as follows. Remember you can always use a difference of two squares to help you to determine the two roots as shown here:
\begin{gathered} k^2-8<0 \\ (k-\sqrt{8})(k+\sqrt{8})<0 \end{gathered}Sketching the curve then gives us:
Using the rules for quadratic inequalities we can see that the final answer is:
-\sqrt{8}<k<\sqrt{8}
Once again the keywords here are “no real roots” and this means that b^2-4 a c<0
For this question we have a=k \quad b=6 k \quad c=2 which will then give us the following:
\begin{aligned} b^2-4 a c & <0 \\ (6 k)^2-4(k)(2) & <0 \\ 36 k^2-8 k & <0 \\ 4 k(9 k-2) & <0 \end{aligned}It is always a good idea to sketch the quadratic curve as this reduces the risk of getting the direction of the inequality symbol the wrong way round.
You can now complete the solution as 0<k<2 / 9
The Discriminant – Harder Exam Questions 1
This is worth a total of 7 marks and so there is going to be a reasonable amount of work involved for this question. Remember that it is very important that you are showing all of your working out in order to achieve maximum marks.
Going back to the question we have the words “two distinct real roots” which means that b^2-4 a c>0
For this particular question we have the following values for a, b and c:
a=1 \quad b=n+1 \quad c=3-3 nSubstituting these into the discriminant will give us another quadratic in terms of n, which can be seen as follows:
\begin{aligned} (n+1)^2-4(1)(3-3 n) & >0 \\ (n+1)(n+1)-4(3-3 n) & >0 \\ n^2+n+n+1-12+12 n & >0 \\ n^2+14 n-11 & >0 \end{aligned}For this particular quadratic you can use the quadratic formula to help you determine the values of n:
\begin{aligned} & a=1 \quad b=14 \quad c=-11 \\ & n=\frac{-(14) \pm \sqrt{(14)^2-4(1)(-11)}}{2(1)} \\ & n=-7+2 \sqrt{15} \quad \text { or } n=-7-2 \sqrt{15} \end{aligned}If you obtain answers as surds it is always best to leave the answer in exact form unless stated otherwise. To help understand how the solutions can be obtained, a sketch of the curve is always useful and this is shown below:
You can then write the final solution as follows:
n<-7-2 \sqrt{15} \quad n>-7+2 \sqrt{15}The Discriminant – Harder Exam Questions 2
For part a) you need to obtain the answer that has been provided in the question. You are not allowed to use it.
Here you need to again understand the wording of the question. The key terms are “does not intersect”. If this is the case then there are no solutions and so b^2-4 a c<0
Given that you have a straight line and a curve you need to initially do simultaneous equations and since both equations are equal to y, you can simply equate the two as shown below:
\begin{aligned} p x^2-4 p x-5 p & =2 x-12 \\ p x^2-4 p x-2 x-5 p+12 & =0 \\ p x^2-(4 p+2) x+(12-5 p) & =0 \end{aligned}You can see that a quadratic has been obtained in terms of “p” and the discriminant can now be used. For this question we have:
a=p \quad b=-(4 p+2) \quad c=(12-5 p)And so:
\begin{aligned} (4 p+2)(4 p+2)-4(p)(12-5 p) & <0 \\ 16 p^2+8 p+8 p+4-4 p(12-5 p) & <0 \\ 16 p^2+16 p+4-48 p+20 p^2 & <0 \\ 36 p^2-32 p+4 & <0 \\ 9 p^2-8 p+1 & <0 \end{aligned}You can see that the term required has been obtained. You will notice that there is quite a bit of algebra involved within the question, so it is important that you are writing clearly and giving yourself plenty of room to do your work.
Again, the quadratic does not factorise so using the quadratic formula gives:
\begin{aligned} & p=\frac{-(-8) \pm \sqrt{(-8)^2-4(9)(1)}}{2(9)} \\ & p=\frac{4+\sqrt{7}}{9} \\ & p=\frac{4-\sqrt{7}}{9} \end{aligned}Sketching a quadratic curve gives:
You can now give solutions as:
\frac{4-\sqrt{7}}{9}<p<\frac{4+\sqrt{7}}{9}The Discriminant – Harder Exam Questions 3
Once again it is a case of understanding the wording of the question. You are told that a straight line is a tangent to a circle. If this is the case then that means that the straight line only makes contact with the circle at one distinct point. So there is just one solution. This means that b^2-4 a c=0
For this particular question you need to first do simultaneous equations by substituting the simpler equation into the equation of the circle and this is shown as follows:
\begin{aligned} & x^2+6 x+(m x-2)^2-8(m x-2)+5=0 \\ & x^2+6 x+(m x-2)(m x-2)-8 m x+16+5=0 \\ & x^2+6 x+m^2 x^2-2 m x-2 m x+4-8 m x+21=0 \\ & \left(m^2+1\right) x^2+(6-12 m) x+25=0 \\ & a=m^2+1 \quad b=6-12 m \quad c=25 \\ & (6-12 m)^2-4\left(m^2+1\right)(25)=0 \\ & (6-12 m)(6-12 m)-100\left(m^2+1\right)=0 \\ & 36-72 m-72 m+144 m^2-100 m^2-100=0 \\ & 44 m^2-144 m-64=0 \\ & 11 m^2-36 m-16=0 \end{aligned}You should study the differences between this question and the previous. They are essentially the same question, but the difference here is that the question has not asked you to obtain a quadratic expression in terms of “m”.
Using the quadratic formula we can know obtain the possible values for “m” as shown below:
\begin{aligned} & m=\frac{-(-36) \pm \sqrt{(-36)^2-4(11)(-16)}}{2(11)} \\ & m=\frac{18+10 \sqrt{5}}{12} \quad m=\frac{18-10 \sqrt{5}}{11} \end{aligned}Questions that involve the discriminant can be quite algebraic intense, but it is first important to be able to translate the wording into the correct notation for maths. The questions do not give you this information directly.
Keep practising these types of questions. You have a good mix of discriminant questions here that show you the detailed steps that are needed. Also remember to sketch the quadratic curve since in most cases, you are ultimately dealing with a quadratic inequality.
If you need additional help with this topic area or indeed any other at A Level, you can use one of our expert online maths tutors, who can steer you to success!