Salient Features of India constitution especially power and functions
C2 st lecture 2 handout
1. Lecture 2 - Quadratic Equations
and Straight Lines
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon Craik
l.stringer@uea.ac.uk s.craik@uea.ac.uk
INTO City/UEA London
2. Lecture 2 skills
factorize a quadratic equation
complete the square
use the quadratic formula
simplify a surd (using a calculator if necessary)
sketch a straight line, given the equation
find the equation of a line, given the gradient and a point
find the equation of a line, given two points
find the midpoint of two points
find the distance between two points
4. Quadratic equations
A quadratic equation is an equation that involves a variable
squared, for example
x2
+ 4x + 3 = 0
3x2
+ 12x − 1 = 0
We have three methods for solving quadratic equations
Factorising
Completing the square
Using the quadratic formula
You need to be able to use ALL THREE METHODS
5. Solving equations equal to zero
If a × b = 0 then a = 0 or b = 0.
For example
If 5x = 0 then x = 0.
If 5(x − 3) = 0 then (x − 3) = 0, so x = 3.
If (x − 3)(x + 4) = 0 then (x − 3) = 0 or (x + 4) = 0.
It follows that x = 3 or x = −4.
6. Factorising when the coefficient of x2
is 1
Change the quadratic equation
x2
+ bx + c = 0 into the form (x + b1)(x + b2) = 0
where b1 and b2 are numbers such that
b1 + b2 = b and b1 × b2 = c
Since
(x + b1)(x + b2) = x2
+ b2x + b1x + b1b2
= x2
+ x(b2 + b2) + b1b2
= x2
+ bx + c
If (x + b1)(x + b2) = 0, then (x + b1) = 0 or (x + b2) = 0.
It follows that x = −b1 or x = −b2
7. Factorising a quadratic equation - example 1
Question: Solve by factorising
x2
+ 8x + 7 = 0
Answer: Find b1 and b2 such that
b1 + b2 = 8 and b1 × b2 = 7
Clearly b1, b2 = 7, 1
(x + 7)(x + 1) = 0
Either (x + 7) = 0 or (x + 1) = 0, so x = −7 and x = −1.
Substitute our solutions into our original equation to check.
(−7)2 + (8 × (−7)) + 7 = 0.
(−1)2 + (8 × (−1)) + 7 = 0
8. Factorising when the coefficient of x2
is not 1
To solve
ax2
+ bx + c = 0
find two numbers b1 and b2 such that
b1 + b2 = b and b1 × b2 = a × c
then split the middle term
ax2
+ bx + c = ax2
+ b1x + b2x + c
and finally factorise the first two terms and the last two
terms (see example 2)
9. Factorising a quadratic equation - example 2
Question: Solve by factorising
6x2
+ 19x + 10 = 0
Answer: Find b1, b2 such that
b1 + b2 = 19 and b1 × b2 = 6 × 10 = 60
Clearly b1, b2 = 15, 4
First split the middle term
6x2
+ 19x + 10 = 6x2
+ 15x + 4x + 10
Now factorise the first two terms and the last two terms
= 3x(2x + 5) + 2(2x + 5) = (3x + 2)(2x + 5)
So (3x + 2)(2x + 5) = 0
It follows that (3x + 2) = 0 or (2x + 5) = 0.
If (3x + 2) = 0, then x = −2
3
If (2x + 5) = 0, then x = −5
2 .
The solutions are x = −2
and x = −5
.
10. Factorising when the coefficient of x2
is not 1
ALTERNATIVE METHOD
Change a quadratic equation of the form
ax2
+bx+c = 0 into the form (a1x+c1)(a2x+c2) = 0
Find all the pairs of factors of a (call them a1, a2), and all
the pairs of factor of c (call them c1, c2).
Experiment to see which combination a1c2 + a2c1 = b.
Solve
15x2
− 14x − 8 = 0
12. Completing the square, x2
+ bx + c = 0
Check the coefficient of x2 is 1: x2 + 6x + 2 = 0
Move c to the right side: x2 + 6x = −2
Add (b
2 )2 to both sides: x2 + 6x + (6
2)2 = −2 + (6
2 )2
Tidy up both sides: x2 + 6x + 9 = 7
Write the left side as (x + b
2 )2: (x + 3)2 = 7
Take the square root of both sides: x + 3 = ±
√
7
Move the constant to the right side: x = −3 ±
√
7
13. Completing the square - example 2
Question: Solve 4x2 − 2x − 3 = 0 by completing the square
Answer: First divide by 4.
x2
−
1
2
x −
3
4
= 0
Check the coefficient of x2 is 1: x2 − 1
2 x − 3
4 = 0
Move c to the right side: x2 − 1
2 x = 3
4
Add (b
2 )2 to both sides: x2 − 1
2 x + (−1
4 )2 = 3
4 + (−1
4)2
Tidy up both sides: x2 − 1
2 x + 1
16 = 13
16
Write the left side as (x + b
2 )2: (x − 1
4 )2 = 13
16
Take the square root of both sides: x − 1
4 = ± 13
16
Move the constant to the right side: x = 1
4 ± 13
16
The solution is x = 1±
√
13
4 .
14. Quadratic formula
To solve ax2 + bx + c = 0, you will be given the quadratic
formula
x =
−b ±
√
b2 − 4ac
2a
Question: Solve 5x2 + 9x − 3 = 0 using the quadratic
formula
Solution:
x =
−9 ± 92 − 4 × 5 × (−3)
2 × 5
So x = −9±
√
141
10 (surd form)
= 0.29, −2.09 (to 2 d.p.)
15. Complete the square to prove the quadratic formula
Take a general quadratic equation ax2 + bx + c = 0.
First we divide through by a, so x2 + b
a x + c
a = 0.
Move the constant term to the right x2 + b
a x = −c
a .
Add ( b
2a )2 to both sides x2 + b
a x + ( b
2a )2 = −c
a + ( b
2a )2.
Write the left side as a square (x + b
2a )2 = −c
a + ( b
2a )2.
Rearrange the right side to get (x + b
2a )2 = b2−4ac
4a2 .
Take the square root of both sides x + b
2a = ±
√
b2−4ac
2a
Then x =
−b±
√
b2−4ac
2a as required.
16. Quadratic formula - example 2
Solve 3x2 − 8x + 2 = 0.
Answer: a = 3, b = −8, c = 2.
By the formula we have
x =
−(−8) ± (−8)2 − 4 × 3 × 2
2 × 3
Simplifying
x =
8 ±
√
64 − 24
6
It follows
x =
8 ±
√
40
6
and so our solutions are
x =
4 +
√
10
3
, x =
4 +
√
10
3
17. Quadratic formula - example 3
Solve x2 = 58x − 2.
Answer:
First rearrange the equation x2 − 58x + 2 = 0
a = 1, b = −58, c = 2. By the formula we have
x =
−(−58) ± (−58)2 − 4 × 1 × 2
2 × 1
Simplifying
x =
58 ±
√
3364 − 8
2
It follows
x =
58 ±
√
3356
2
and so our solutions are
x = 58 −
√
839, x = 58 +
√
839
18. Surds
Surds must be presented in a simplified form.
Your calculator will simplify surds for you!
There must be no square factor under the square root
sign. To simplify use the rule
√
a × b =
√
a ×
√
b
Simplify
√
50√
50 =
√
2 × 25 =
√
2 ×
√
25 = 5
√
2.
√
10 can not be simplified.
20. The gradient of a straight line
Gradient (slope) =
Vertical change
Horizontal change
=
Change in y
Change in x
=
∆y
∆x
=
Rise
Run
∆x (run)
∆y (rise)
x
y
For each step to the right, the gradient tells you how many
steps up (or down)
21. The gradient of a straight line
−4 −2 2 4
−4
−2
2
4
x
y
−4 −2 2 4
−4
−2
2
4
x
y
−4 −2 2 4
−4
−2
2
4
x
y
−4 −2 2 4
−4
−2
2
4
x
y
22. A straight line - the equation, gradient and y-intercept
A straight line has equation
y = mx + c
c is the y-intercept. This is the y-coordinate of the point
where the line passes through the y-axis. The line
intercepts the y-axis at the point (0, c).
m is the gradient of the line - for every one step to the right,
you go m steps upwards
A steep line has a large gradient (m > 1 or m < −1.)
A shallow line has a small gradient (−1 < m < −1.)
A horizontal line has gradient 0. In this case the equation
of the line is y = c.
A vertical line has gradient ∞. In this case the equation of
the line is x = a (for some constant a.)
23. Sketch a line with equation y = x + 2
−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
•
•
x
y
Find the y-intercept
Let x = 0
Then y = 0 + 2 = 2
The line crosses the y-axis
at (0, 2)
Find the x-intercept
Let y = 0
Then 0 = x + 2, so x = −2
The line crosses the x-axis
at (−2, 0)
Plot the intercepts
Check the gradient
Sketch the line
24. Sketch a line
To sketch the line with equation
y = mx + c
Find the coordinates of two points on the line - usually it is
best to find the y-intercept and x -intercept.
Plot the two points.
Check the gradient.
Sketch the line
25. A line defined by a gradient and a point
A straight line can also be defined by a gradient m and a
point (x0, y0) through which the line passes.
To get the equation of the line substitute x0 and y0 into the
formula y = mx + c to find c.
Question: Find the equation of the straight line with
gradient 0.5 that passes through (6, 2).
Answer: Substitute 2 = 0.5 × 6 + c. So 2 = 3 + c and
c = −1. The equation is
y = 0.5x − 1
or equivalently
y =
1
2
x − 1
26. A line defined by two points
A straight line can also be defined by 2 points (x0, y0) and
(x1, y1) through which the line passes.
To get the equation of the line we need to find the gradient
m and the y-intercept c.
The gradient is given by the formula:
m =
y1 − y0
x1 − x0
To find c we substitute in one of our points, say (x0, y0) into
the line equation y = mx + c.
27. A line defined by two points - example
Question: Find the equation of the straight line passing
through the points (1, 1) and (−1, 3).
Answer: First find the gradient
m =
3 − 1
−1 − 1
=
2
−2
= −1
Substitute one of the points into the line equation to find c.
We have 1 = −1 × 1 + c so c = 2.
The equation of our line is
y = −x + 2.
We should now check our answer. Substitute in the other
point (−1) × (−1) + 2 = 3.
28. Midpoint
To find the mid-point of two points (x0, y0) and (x1, y1) we
calculate the midpoint of x0 and x1 and the midpoint of y0
and y1.
The midpoint is
x0 + x1
2
,
y0 + y1
2
Question: What is the midpoint of the points (1, 1) and
(−1, 3)?
Answer: The midpoint is
1 + (−1)
2
,
1 + 3
2
= (0, 2)
29. Distance
To find the distance between two points (x0, y0) and
(x1, y1) we use the formula
(x1 − x0)2 + (y1 − y0)2
Question: What is the distance from (1, 1) to (−1, 3)?
Answer: The distance is
((−1) − 1)2 + (3 − 1)2 = (−2)2 + 22 =
√
8 = 2
√
2
