QUADRATICS
1. Quadratic Equations
A quadratic equation is one of the form . There are three methods of solution.
a) Factorising
Example 1 : Solve .
We first rearrange so that 0 is on the RHS.
We say that −7 and 2 are the roots (solutions) of the equation.
Example 2 : Solve .
Sometimes we can get away without rearranging to get 0 on the RHS.
Example 3 : Solve .
Note that the alternative method is…
Example 4 : Solve .
Now try this by factorising and see what happens!
C1 p17 Ex 2B
Not every quadratic will factorise, however; in these cases we must use one of the two following methods.
b) Completing the Square
Consider the quadratic expression . We can represent the part by this diagram.
We can rearrange this rectangle into a square with a corner missing...
The area of the rectangle can now be written as
The is the area of the entire square, and the −9 is the area of the missing corner.
So can be rewritten as
This process is known as completing the square. It is a very useful technique when dealing with quadratics, due to the fact that the x now only appears once in the expression instead of twice. In general an expression of the form becomes .
Example 3 : Solve the equation .
Example 4 : Solve the equation .
Example 5 : Solve the equation .
We require the coefficient of to be 1. We therefore divide the equation through by 2.
Example 6 : Solve the equation .
C1 p18 Ex 2C, p19 Ex 2D
c) Quadratic Roots Formula
The roots of the equation are given by
The formula can be proved by completing the square.
Example 7 : Solve the equation .
Example 8 : Solve the equation .
Example 9 : Solve the equation .
We cannot take the square root of a negative number, and so the equation has no solutions.
From examples 7, 8 and 9, we see that a quadratic equation can have two, one or no solutions. To understand this, consider again the quadratic roots formula.
We call the discriminant, and it determines how many solutions an equation has.
•If, we have two solutions (see example 7), because one solution will involve and the other .
•If, we have one solution (see example 8), because and are both 0, and therefore the ‘two solutions’ will be exactly the same. Because of this, the single solution is sometimes called a ‘repeated root’ or ‘coincident root’
•If, we have no solutions (see example 9), because we cannot take the square root of a negative number.
We can also see why a quadratic will only factorise if the discriminant is a square number. Only then will the roots be rational.
C1 p21 Ex 2E
2. Sketching Quadratic Functions
A quadratic function , has a characteristic U-shape. When sketching quadratics we must consider four things.
1. Whether the graph is ‘right-way-up’ () or ‘upside-down’ ().
2. Where the graph crosses the x-axis.
3. Where the graph crosses the y-axis.
4. The coordinates of the maximum/minimum point (use completing the square).
Example 1 : Sketch the graph of
1. The coefficient of is positive, and so the graph is the ‘right-way-up’.
2. The graph crosses the x-axis when . Therefore
So the graph crosses the x-axis at (−7, 0) and (5, 0).
3. The graph crosses the y-axis when . So the crossing point is (0, −35).
4. To find the minimum point of the graph, we complete the square.
Now is either zero (when) or positive (when x is anything else). Therefore y will have a minimum value when. We see that this minimum value is −36. So the coordinates of the minimum point are (−1, −36).
Compiling all our information onto one sketch...
Notice the symmetry about the line . Exam questions often ask for the co-ordinates of intercepts with the axes, so write (5, 0) on the x-axis, and not just 5 (which is an ordinate).
Activity1:This can be checked on a graphic calculator.
•Select GRAPH (item 5 in the menu) and type the function in for Y1 (press X, θ, T followed by + 2 X, θ, T − 35 EXE).
•Set the range by selecting V-WIN (shift F3) and entering appropriate values.
•Select DRAW (F6).
•Find the minimum by selecting G-SOLV → MIN (shift F5 F3).
•The y-intercept can be found by doing G-SOLV → Y-ICPT, and the x-intercepts by doing G-SOLV → ROOT (press the right arrow to move from the first to the second root).
On Autograph, draw the curve and select it by clicking on the pointer icon and then the curve. Open the results box, and then from the Object menu select ‘solve ’ for the roots and ‘solve ’ for the turning point. The contents of the results box can be deleted at any time if you find them annoying.
Example 2 : Sketch the graph of .
1. The coefficient of is negative, and so the graph is the ‘upside-down.'
2. The graph crosses the x-axis when . Therefore
So the graph crosses the x-axis at (4.345, 0) and (−0.345, 0).
3. The graph crosses the y-axis when . So the crossing point is (0, 3).
4.To complete the square, we take a factor of −2 outside the part of the equation involving x, to reduce the coefficient of to 1. (We cannot divide by −2, because we are manipulating an expression, not solving an equation).
Now is either zero (when ) or negative (when x is anything else). Therefore y will have a maximum value when x = 2. We see that this maximum value is 11. So the graph has a maximum point at (2, 11).
Compiling all our information onto one sketch...
Recommended additional examples : , .
C1 p22 Ex 2F Q1
3. Quadratic Graphs and the Discriminant
We saw in section 1 that the discriminant determines the number of roots of a quadratic. In section 2, we saw that the roots of a quadratic are the points where the graph crosses the x-axis. So the discriminant determines how many times the graph crosses the x-axis.
2 roots 1 root 0 roots
The two roots are called distinct roots, whilst the one root is known as a repeated or coincident root. (Imagine the quadratic graph slowly rising so that the two roots eventually coincide to make a single root).
Activity2:Check this by drawing the graphs of the three quadratic equations at the end of section 1.
Example 1 : If has a repeated root, find the value of a.
For a repeated root,
Example 2 : Find the range of values of k if the equation has distinct roots.
For distinct roots, we require
Activity3:Both of these answers can be checked on graphic calculators or Autograph. With Autograph in Example 2, type in , and animate as k moves from –3 to 3.
Example3:Prove that the line is a tangent to the curve , and find the point of contact.
There are three possibilities with a straight line and a quadratic : two points of intersection, one (in which case we have a tangent) or none.
We begin by assuming there are points of intersection. At these points,
We see that there is only one solution, and so the line is tangent to the curve, touching at the point (1, –3). We get the y-coordinate by substituting into the equation of either the straight line or the quadratic.
C1 p22 Ex 2F Q2,3, p23 Ex 2G
4. Linear Inequalities
We solve these in exactly the same way as linear equations, with one exception : an inequality becomes false if we multiply or divide by a negative number. For example,
Multiplying by –2...
...which is clearly false. So if we multiply or divide by a negative number, we must reverse the direction of the inequality sign.
Example 1 : Solve . / Example 2 : Solve .C1 p31 Ex 3D (should not be necessary however)
5. Quadratic Inequalities
These should always be solved by means of a sketch graph.
Example 1 : Solve the inequality .
We now sketch the graph of and consider the portion above the x-axis.
Example 2 : Solve the inequality .
This time we want the portion of the graph on or below the x-axis.
Example 3 : Solve the inequality .
This time we want the portion of the graph below the x-axis.
Example 4 : Find the values of x for which and .
We also have
Combining the two inequalities, we have our final solution
Example 5 : A rectangle has a perimeter of 40cm. Find the range of possible values of the length of the rectangle if the area is to be less than 75cm2.
Let the length be x and the width y. We have
We require
However, we also have the condition that . Combining with the above solution,
Example6:A farmer has 60 metres of fencing with which to form a rectangular enclosure, with one side taken up by a long stone wall. She wishes to enclose an area of at least 410m2. What is the range of possible values for the length of the enclosure?
We have
So we require
We now sketch the graph of. As this does not factorise, we use the quadratic roots formula to find the points of intersection with the x-axis.
We see that the solution to the inequality is.
C1 p35 Ex 3E Topic Review : Expanding, Factorising and Quadratics