Lecture 2 - Quadratic Equationsand Straight Lines

C2 Foundation Mathematics (Standard Track)

Dr Linda Stringer Dr Simon Craikl.stringer@uea.ac.uk s.craik@uea.ac.uk

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