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 √ square rootI gradientI interceptI >
I ≥I <
I ≤I ∞
Quadratic equations
I A quadratic equation is an equation that involves a variablesquared, for example
x2 + 4x + 3 = 0
3x2 + 12x − 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 a× b = 0 then a = 0 or b = 0.
For exampleI If 5x = 0 then x = 0.I If 5(x − 3) = 0 then (x − 3) = 0, so x = 3.I If (x − 3)(x + 4) = 0 then (x − 3) = 0 or (x + 4) = 0.
It follows that x = 3 or x = −4.
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 × 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 = −b1 or x = −b2
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 × 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.I Substitute our solutions into our original equation to check.
(−7)2 + (8× (−7)) + 7 = 0.(−1)2 + (8× (−1)) + 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 × 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 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 × b2 = 6× 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 = −2
3If (2x + 5) = 0, then x = −5
2 .I The solutions are x = −2
3 and x = −52 .
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 − 14x − 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 = −2
Add (b2 )2 to both sides: x2 + 6x + (6
2)2 = −2 + (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 = ±√
7
Move the constant to the right side: x = −3±√
7
Completing the square - example 2
Question: Solve 4x2 − 2x − 3 = 0 by completing the squareAnswer: First divide by 4.
x2 − 12
x − 34
= 0
Check the coefficient of x2 is 1: x2 − 12x − 3
4 = 0Move c to the right side: x2 − 1
2x = 34
Add (b2 )2 to both sides: x2 − 1
2x + (−14)2 = 3
4 + (−14)2
Tidy up both sides: x2 − 12x + 1
16 = 1316
Write the left side as (x + b2 )2: (x − 1
4)2 = 1316
Take the square root of both sides: x − 14 = ±
√1316
Move the constant to the right side: x = 14 ±
√1316
The solution is x = 1±√
134 .
Quadratic formula
I To solve ax2 + bx + c = 0, you will be given the quadraticformula
x =−b ±
√b2 − 4ac
2a
I Question: Solve 5x2 + 9x − 3 = 0 using the quadraticformula
I Solution:
x =−9±
√92 − 4× 5× (−3)
2× 5
So x = −9±√
14110 (surd form)
= 0.29,−2.09 (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 = − c
a .
I Add ( b2a )2 to both sides x2 + b
a x + ( b2a )2 = − c
a + ( b2a )2.
I Write the left side as a square (x + b2a )2 = − c
a + ( b2a )2.
I Rearrange the right side to get (x + b2a )2 = b2−4ac
4a2 .
I Take the square root of both sides x + b2a = ±
√b2−4ac
2a
I Then x =−b±√
b2−4ac2a as required.
Quadratic formula - example 2I Solve 3x2 − 8x + 2 = 0.I Answer: a = 3, b = −8, c = 2.I By the formula we have
x =−(−8)±
√(−8)2 − 4× 3× 22× 3
I Simplifying
x =8±√
64− 246
I It follows
x =8±√
406
I and so our solutions are
x =4 +√
103
, x =4 +√
103
Quadratic formula - example 3I Solve x2 = 58x − 2.I Answer:
First rearrange the equation x2 − 58x + 2 = 0a = 1, b = −58, c = 2. By the formula we have
x =−(−58)±
√(−58)2 − 4× 1× 22× 1
Simplifying
x =58±
√3364− 82
It follows
x =58±
√3356
2and so our solutions are
x = 58−√
839, x = 58 +√
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 rule√
a× b =√
a×√
b
I Simplify√
50√50 =
√2× 25 =
√2×√
25 = 5√
2.I√
10 can not be simplified.
Cartesian coordinates, (x, y)
−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
•(4,2)
•
•
•
x
y
The gradient of a straight line
Gradient (slope) =Vertical change
Horizontal change=
Change in yChange in x
=∆y∆x
=RiseRun
∆x (run)
∆y (rise)
x
y
For each step to the right, the gradient tells you how manysteps up (or down)
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
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 < −1.)I A shallow line has a small gradient (−1 < m < −1.)I A horizontal line has gradient 0. In this case the equation
of the line is y = c.I A vertical line has gradient∞. In this case the equation of
the line is x = a (for some constant a.)
Sketch a line with equation y = x + 2
−6 −4 −2 2 4 6
−6
−4
−2
2
4
6
•
• 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 = −2The line crosses the x-axisat (−2,0)
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 = 0.5× 6 + c. So 2 = 3 + c andc = −1. The equation is
y = 0.5x − 1
or equivalently
y =12
x − 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 − y0
x1 − 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 (−1,3).
I Answer: First find the gradient
m =3− 1−1− 1
=2−2
= −1
I Substitute one of the points into the line equation to find c.We have 1 = −1× 1 + c so c = 2.
I The equation of our line is
y = −x + 2.
I We should now check our answer. Substitute in the otherpoint (−1)× (−1) + 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(−1,3)?
I Answer: The midpoint is(1 + (−1)
2,1 + 3
2
)= (0,2)
Distance
I To find the distance between two points (x0, y0) and(x1, y1) we use the formula√
(x1 − x0)2 + (y1 − y0)2
I Question: What is the distance from (1,1) to (−1,3)?I Answer: The distance is√
((−1)− 1)2 + (3− 1)2 =√
(−2)2 + 22 =√
8 = 2√
2