Core 1 revision - Equations and inequalities

When you are given an equation in the form of ax²+bx+c=0 it is called a quadratic equation. If you are given an equation in this form, but it isn't equal to 0, you can set it equal to 0 to work it out.

To find a solution to this, you need to put it into double brackets, so that when expanded the brackets should form the original equation. The numbers in the double brackets need to multiply to give c and these two numbers should add together to give b.

If a is greater than 1, you need to remember that when finding b, the number is going to be greater than it would normally be. For example, the quadratic equation of 3x²+13x-10=0 could provide some difficulty. You need the two numbers to multiply to give -10 (notice the sign) but they need to add together to give 13, but when looking at this, you need to consider the 3x²

When put into double brackets this would give (3x-2)(x+5) and when you expand the brackets you would get b=13 so you have to make sure you look at the value of a when working this out as it could give a different answer than required.

Another way to factorise is completing the square. If you are given two numbers that are square numbers (like 4x²+25=0) you can put it into brackets using the square roots of both with opposing symbols - (2x-5)(2x+5).

Another example would be having two different 'letter' values. Like you do with numbers, you would put it into double brackets and instead of putting numbers in to get the values, you would use the appropriate letters making sure that when you expand the brackets your original equation is still correct.

When you want to find a solution to a quadratic equation, at least one of the double brackets must equal 0 as a quadratic equation is ALWAYS equal to 0. Therefore, in the example (x-6)(x+2)=0, x must be either 6 or -2 as then the brackets would allow this to be correct. However, if you had (2x-5)(x-2), one solution would be 2, as explained above, and the (2x-5) still needs to be equal to 0:

2x-5=0 --> 2x=5 --> x=5/2=2.5

So the two solutions would be 2.5 and 2. But what you need to remember is that you could be given an equation that doesn't =0, therefore, you need to make it equal to 0 before moving on to the next steps as I explained before.

You could get given an exam question that wants you to set up a quadratic equation and solve it.

An example would be: There is a field with an area of 525m(squared) with a width of x+15 and a height of x-5. What is the value of x?

First you would set up the equation - (x+15)(x-5)=525 as the two brackets multiply to give you the overall area.

Then: (x+15)(x-5)=525

x²-5x+15x-75=525

x²+10x-600=0

(x-20)(x+30)=0

So x is 20 or -30 - however, x can't be two values in this case so you need to work out which one it is. So, you can plug it back into one of the first equations. 

x+15 --> 20+15=35 - this could work

x+15 --> -30+15=-15 - this can't work as you can't have a negative value so x is 20.

x= -b ±√b²-4ac

           2a 

This is the quadratic formula. With each letter associating with its respective letter from ax²+bx+c=0.

There is a third way of solving quadratic equations, completing the square.

For this you need an equation that has two forms of square numbers. For example, x²+25 has two forms of sqaure numbers in it, this means we can use completing the square. To do this, you put the square root of the equation in brackets and sqaure the whole bracket, so for this example, it would  be (x+5)².

Finding the intersection point when you have two equations, one quadratic and one linear.

1. First plot the quadratic equation on a graph.
2. Now plot the linear equation - you can do this by making y=0 and x=0 to find the different values of where they intercept the axis.
3. You should then be able to find the two points visually where they cross.

To make sure that this is correct, you can substitute the values you were given back into the two equations.

For example: If your two equations are y=x² +2x-4 and y=x+2.

After drawing the graph (which I can't put on here) the points will intersect (2,4) and (-3, -1). If you substitute the x value into the both equations, you should get the y-value you started with and the same for the second value of points.

There is another way of doing this that doesn't involve drawing any graphs which can get a bit frustrating!

When you are given two equations which are both erqual to y, you can make them equal to each other. This will then give you a quadratic equation which will then give you two x values from the double brackets. However, the y values still need to be found.

To do this, you can pick one of the equations, normally the linear as this will cause the least problems, and substitute the x-value into the equation to give you one y-value. To then find the second y-value, you need to substitue the second x-value into the equation and do the same again.

Exmainers like to phrase the question differently despite them meaning the same thing, just to confuse you and make the exam seem harder. 

The phrases "Solve simultaneously" and "find where the two lines intersect" both mean the same thing so you follow the steps shown previously.

A linear equation could also be given, however, it could form a circle instead of a straight line, which requires something slightly different to begin with.

X² + y² = 4 If you are given an equation like this, it will form a circle. When it forms a circle, you need to remember that, each form can have a positive or a negative value as a negative times a negative gives a positive. (+/-) x + (+/-) y = (+/-)√4

So, in this example, when x=0, y can be + or - 2, and when y=0, x can be + or - 2. When joined this will form a circle around the point (0,0).

This is how to form the circle, however, you still need to find where the two equations intersect.

So, using the graph method, you draw the linear equation on the same graph as the circle and read off the values where the two equations meet.

Unlike the first example, you now have two equations where there is no clear y= making it slightly harder to combine the two equations. One equation will be given to you in the form of y= and the first equation will have a y in it, therefore, you can use the substitution method. 

x² +y² = 4

y = x+2

If you are given these two equations, you can substitute the second equation into the first squaring the (x+2). After doing this, you will get a quadratic equation, which you need to make into the form of 0 in order to solve. It may be that you don't factorise it into double brackets, but either way you will get two answers for x.

Like before, you then substitute these values into an equation (in this example the easier one would be y=x+2) and find the values for y.

That was a more simple example of a circle equation another one would be:

(x-2)² + (y-4)² = 25

If you set x and y to 0 you get the value of (2,4). This is the centre point of the circle and we are told the radius through the square root of 25, which is 5.

If you are then given the equation y=-x+7, you follow the method mentioned previously and draw the linear equation onto the graph. After reading off the points, for this example, the solutions are (-1,8) and (6,1).

Like the other examples, you can solve this without a graph and use the substitution method. By substituting the values, you can tidy it up and then solve the quadratic equation and resubstitute these values to get the y-values.

If it comes in a different form (eg. x+y=9) rearrange it to get what y is and work your way through the question just like before.

These now get complicated.

When given in the form xy=3, it can be rewritten as y=3/x which is a reciprocol graph. - drawing a graph may help more with this. Now find the intersection points of the linear equation.

However, it can be done without drawing a graph. Instead of turning the equation into a reciprocol, substitute the value of x or y (whichever is given) into the second equation and find the values, substituting these back depending on which you find.

xy=3

x-y=2 --> y=x-2

x(x-2)=3 --> x² -2x=3 --> x² -2x-3=0 --> (x-3)(x+1) so x=3 or -1

x-y=2-->3-y=2-->y=1 OR x-y=2-->-1-y=2-->y=-3

(3,1) and (-1,-3)

Sometimes, you can be given two equations, one which is a circle and one which is a quadratic. In this example, the two equations are x²+y²=16 and y=x² +1.

If you prefer drawing the graph to find the intersection points, there is an extra step included as it is more than likely that your answer won't be a whole number. with the example above, you have a common factor, x²

With this, you can find your solutions. By making the x squared the equation of the quadratic equation, you can then substitute it back into the first equation like this:

x²+y²=16 and y=x² +1.

x²=y-1

y-1 +y²=16 ày²-y-17=0 à you cannot factorise this, so instead you can use the quadratic formula. This gives you y=3.7 or -4.7. Then going back to the quadratic equation, you can find the values of x as you have the values for y. Your answer that you then get is x= +/- 1.6.

(1.6,3.7) or (-1.6,3.7). However, this is only for the y value of 3.7, so what happens with -4.7?

The result you would get is ±√-5.7 which is impossible so the two solutions you had above are the answer.

As well as being given a linear equation that you need to solve using a graph, you could be given a linear equation in the form of an inequality.

A key thing that you need to remember with inequalities is that if you ever multiply or divide by a negative number, the symbol needs to turn around.

If you were given the two inequalities of 4x+7>3 and 17<11+2x you would find the values of x to be x>-1 and x>3 correspondingly. From this your could plot them on a number line to find the values for x where the inqualities will both be true.

However, if you are given a quadratic inequality you need to:

1) Solve the corresponding quadratic equation

2) Then sketch the graph of the quadratic function (formed from the inequality)

3) Then use this graph ti find the required set of value. - (this is generally the area around the line that does not form part of the solution.)