c2 st lecture 2 handout
TRANSCRIPT
Lecture 2 - Quadratic Equationsand Straight Lines
C2 Foundation Mathematics (Standard Track)
Dr Linda Stringer Dr Simon [email protected] [email protected]
INTO City/UEA London
Lecture 2 skills
I factorize a quadratic equationI complete the squareI use the quadratic formulaI simplify a surd (using a calculator if necessary)
I sketch a straight line, given the equationI find the equation of a line, given the gradient and a pointI find the equation of a line, given two pointsI find the midpoint of two pointsI find the distance between two points
Lecture 2 vocabulary and symbols
I quadraticI surdI β square rootI gradientI interceptI >
I β₯I <
I β€I β
Quadratic equations
I A quadratic equation is an equation that involves a variablesquared, for example
x2 + 4x + 3 = 0
3x2 + 12x β 1 = 0
I We have three methods for solving quadratic equations
FactorisingCompleting the square
Using the quadratic formula
You need to be able to use ALL THREE METHODS
Solving equations equal to zero
If aΓ b = 0 then a = 0 or b = 0.
For exampleI If 5x = 0 then x = 0.I If 5(x β 3) = 0 then (x β 3) = 0, so x = 3.I If (x β 3)(x + 4) = 0 then (x β 3) = 0 or (x + 4) = 0.
It follows that x = 3 or x = β4.
Factorising when the coefficient of x2 is 1
I 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 Γ b2 = c
I Since
(x + b1)(x + b2) = x2 + b2x + b1x + b1b2
= x2 + x(b2 + b2) + b1b2
= x2 + bx + c
I If (x + b1)(x + b2) = 0, then (x + b1) = 0 or (x + b2) = 0.It follows that x = βb1 or x = βb2
Factorising a quadratic equation - example 1
I Question: Solve by factorising
x2 + 8x + 7 = 0
I Answer: Find b1 and b2 such that
b1 + b2 = 8 and b1 Γ b2 = 7
Clearly b1,b2 = 7,1
(x + 7)(x + 1) = 0
Either (x + 7) = 0 or (x + 1) = 0, so x = β7 and x = β1.I Substitute our solutions into our original equation to check.
(β7)2 + (8Γ (β7)) + 7 = 0.(β1)2 + (8Γ (β1)) + 7 = 0
Factorising when the coefficient of x2 is not 1
I To solveax2 + bx + c = 0
find two numbers b1 and b2 such that
b1 + b2 = b and b1 Γ b2 = aΓ c
then split the middle term
ax2 + bx + c = ax2 + b1x + b2x + c
and finally factorise the first two terms and the last twoterms (see example 2)
Factorising a quadratic equation - example 2I Question: Solve by factorising
6x2 + 19x + 10 = 0
I Answer: Find b1,b2 such that
b1 + b2 = 19 and b1 Γ b2 = 6Γ 10 = 60
Clearly b1,b2 = 15,4First 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) = 0It follows that (3x + 2) = 0 or (2x + 5) = 0.If (3x + 2) = 0, then x = β2
3If (2x + 5) = 0, then x = β5
2 .I The solutions are x = β2
3 and x = β52 .
Factorising when the coefficient of x2 is not 1I ALTERNATIVE METHODI Change a quadratic equation of the form
ax2+bx+c = 0 into the form (a1x+c1)(a2x+c2) = 0
I Find all the pairs of factors of a (call them a1,a2), and allthe pairs of factor of c (call them c1, c2).
I Experiment to see which combination a1c2 + a2c1 = b.I Solve
15x2 β 14x β 8 = 0
Completing the square
Al-Khwarizmi - the father of algebra
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 = β2
Add (b2 )2 to both sides: x2 + 6x + (6
2)2 = β2 + (62)2
Tidy up both sides: x2 + 6x + 9 = 7
Write the left side as (x + b2 )2: (x + 3)2 = 7
Take the square root of both sides: x + 3 = Β±β
7
Move the constant to the right side: x = β3Β±β
7
Completing the square - example 2
Question: Solve 4x2 β 2x β 3 = 0 by completing the squareAnswer: First divide by 4.
x2 β 12
x β 34
= 0
Check the coefficient of x2 is 1: x2 β 12x β 3
4 = 0Move c to the right side: x2 β 1
2x = 34
Add (b2 )2 to both sides: x2 β 1
2x + (β14)2 = 3
4 + (β14)2
Tidy up both sides: x2 β 12x + 1
16 = 1316
Write the left side as (x + b2 )2: (x β 1
4)2 = 1316
Take the square root of both sides: x β 14 = Β±
β1316
Move the constant to the right side: x = 14 Β±
β1316
The solution is x = 1Β±β
134 .
Quadratic formula
I To solve ax2 + bx + c = 0, you will be given the quadraticformula
x =βb Β±
βb2 β 4ac
2a
I Question: Solve 5x2 + 9x β 3 = 0 using the quadraticformula
I Solution:
x =β9Β±
β92 β 4Γ 5Γ (β3)
2Γ 5
So x = β9Β±β
14110 (surd form)
= 0.29,β2.09 (to 2 d.p.)
Complete the square to prove the quadratic formula
I Take a general quadratic equation ax2 + bx + c = 0.I First we divide through by a, so x2 + b
a x + ca = 0.
I Move the constant term to the right x2 + ba x = β c
a .
I Add ( b2a )2 to both sides x2 + b
a x + ( b2a )2 = β c
a + ( b2a )2.
I Write the left side as a square (x + b2a )2 = β c
a + ( b2a )2.
I Rearrange the right side to get (x + b2a )2 = b2β4ac
4a2 .
I Take the square root of both sides x + b2a = Β±
βb2β4ac
2a
I Then x =βbΒ±β
b2β4ac2a as required.
Quadratic formula - example 2I Solve 3x2 β 8x + 2 = 0.I Answer: a = 3, b = β8, c = 2.I By the formula we have
x =β(β8)Β±
β(β8)2 β 4Γ 3Γ 22Γ 3
I Simplifying
x =8Β±β
64β 246
I It follows
x =8Β±β
406
I and so our solutions are
x =4 +β
103
, x =4 +β
103
Quadratic formula - example 3I Solve x2 = 58x β 2.I Answer:
First rearrange the equation x2 β 58x + 2 = 0a = 1, b = β58, c = 2. By the formula we have
x =β(β58)Β±
β(β58)2 β 4Γ 1Γ 22Γ 1
Simplifying
x =58Β±
β3364β 82
It follows
x =58Β±
β3356
2and so our solutions are
x = 58ββ
839, x = 58 +β
839
Surds
I Surds must be presented in a simplified form.I Your calculator will simplify surds for you!I There must be no square factor under the square root
sign. To simplify use the ruleβ
aΓ b =β
aΓβ
b
I Simplifyβ
50β50 =
β2Γ 25 =
β2Γβ
25 = 5β
2.Iβ
10 can not be simplified.
Cartesian coordinates, (x, y)
β6 β4 β2 2 4 6
β6
β4
β2
2
4
6
β’(4,2)
β’
β’
β’
x
y
The gradient of a straight line
Gradient (slope) =Vertical change
Horizontal change=
Change in yChange in x
=βyβx
=RiseRun
βx (run)
βy (rise)
x
y
For each step to the right, the gradient tells you how manysteps up (or down)
The gradient of a straight line
β4 β2 2 4
β4
β2
2
4
x
y
β4 β2 2 4
β4
β2
2
4
x
y
β4 β2 2 4
β4
β2
2
4
x
y
β4 β2 2 4
β4
β2
2
4
x
y
A straight line - the equation, gradient and y-intercept
I A straight line has equation
y = mx + c
I c is the y-intercept. This is the y-coordinate of the pointwhere the line passes through the y-axis. The lineintercepts the y-axis at the point (0, c).
I m is the gradient of the line - for every one step to the right,you go m steps upwards
I A steep line has a large gradient (m > 1 or m < β1.)I A shallow line has a small gradient (β1 < m < β1.)I A horizontal line has gradient 0. In this case the equation
of the line is y = c.I A vertical line has gradientβ. In this case the equation of
the line is x = a (for some constant a.)
Sketch a line with equation y = x + 2
β6 β4 β2 2 4 6
β6
β4
β2
2
4
6
β’
β’ x
y
I Find the y-interceptLet x = 0Then y = 0 + 2 = 2The line crosses the y-axisat (0,2)
I Find the x-interceptLet y = 0Then 0 = x + 2, so x = β2The line crosses the x-axisat (β2,0)
I Plot the interceptsI Check the gradientI Sketch the line
Sketch a line
To sketch the line with equation
y = mx + c
I Find the coordinates of two points on the line - usually it isbest to find the y-intercept and x -intercept.
I Plot the two points.I Check the gradient.I Sketch the line
A line defined by a gradient and a point
I A straight line can also be defined by a gradient m and apoint (x0, y0) through which the line passes.
I To get the equation of the line substitute x0 and y0 into theformula y = mx + c to find c.
I Question: Find the equation of the straight line withgradient 0.5 that passes through (6,2).
I Answer: Substitute 2 = 0.5Γ 6 + c. So 2 = 3 + c andc = β1. The equation is
y = 0.5x β 1
or equivalently
y =12
x β 1
A line defined by two points
I A straight line can also be defined by 2 points (x0, y0) and(x1, y1) through which the line passes.
I To get the equation of the line we need to find the gradientm and the y-intercept c.
I The gradient is given by the formula:
m =y1 β y0
x1 β x0
I To find c we substitute in one of our points, say (x0, y0) intothe line equation y = mx + c.
A line defined by two points - example
I Question: Find the equation of the straight line passingthrough the points (1,1) and (β1,3).
I Answer: First find the gradient
m =3β 1β1β 1
=2β2
= β1
I Substitute one of the points into the line equation to find c.We have 1 = β1Γ 1 + c so c = 2.
I The equation of our line is
y = βx + 2.
I We should now check our answer. Substitute in the otherpoint (β1)Γ (β1) + 2 = 3.
Midpoint
I To find the mid-point of two points (x0, y0) and (x1, y1) wecalculate the midpoint of x0 and x1 and the midpoint of y0
and y1.I The midpoint is (
x0 + x1
2,y0 + y1
2
)
I Question: What is the midpoint of the points (1,1) and(β1,3)?
I Answer: The midpoint is(1 + (β1)
2,1 + 3
2
)= (0,2)
Distance
I To find the distance between two points (x0, y0) and(x1, y1) we use the formulaβ
(x1 β x0)2 + (y1 β y0)2
I Question: What is the distance from (1,1) to (β1,3)?I Answer: The distance isβ
((β1)β 1)2 + (3β 1)2 =β
(β2)2 + 22 =β
8 = 2β
2