quadratics ax 2 + bx + c. multiplying brackets (foil) x 2 +5x +6 outside ( x+ 3)(x + 2) first x2x2...

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QUADRATICS ax 2 + bx + c

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Page 1: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

QUADRATICS

ax2 +

bx

+ c

Page 2: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

MULTIPLYING BRACKETS (FOIL)

x2 +5x +6

Outside

( x+ 3)(x + 2)

First

x2

InsideLast

+2x

+3x

+6

Page 3: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

MULTIPLYING BRACKETS (FOIL) (WITH MINUS NUMBERS)

x2 +2x -8

Outside

( x+ 4)(x - 2)

First

x2

InsideLast

-2x

+4x

-8

Page 4: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

MULTIPLYING BRACKETS (FOIL) (WITH MINUS NUMBERS)

x2 -7x +12

Outside

( x-3)(x - 4)

First

x2

InsideLast

-4x

-3x+12

Page 5: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

y

xy = x2 +3x -4

When y =0

X = 1 or -4

Page 6: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

QUADRATICS

You can find a

quadratic from it’s

roots

Page 7: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

QUADRATICS

For example, x = 3 or -2

When y =0 (x - 3) =0

or (x + 2 ) =0Multiply the

bracketsx2 - x - 6 = 0

Page 8: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

FACTORISING AND SOLVING QUADRATICS

x2 +5x +6

( x )(x )

The numbers have to add up to +5 and multiply to make +6

+3

+2

The x’s have to multiply to make the first term

Page 9: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

CHECK IT OUT

Outside

( x+ 3)(x + 2)

First

Inside

Last

x2 +2x + 3x + 6 = (x2 + 5x +6)

Page 10: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

FACTORISING AND SOLVING QUADRATICS

2x2 - 4x - 6

( 2x )(x )

The numbers have to add up to -4 and multiply to make -6

+2

- 3

The x’s have to multiply to make the first term

2x x -3 = -6x

-6x +2x = -4x

Page 11: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

EXAMPLES TO REMEMBER

(a – b)(a + b)

a2 +ab –ab – b2

= a2 – b2

Page 12: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

EXAMPLES TO REMEMBER

The same applies to all these type

of equations

a2 – 9

= (a-3)(a+3)

Page 13: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

EXAMPLES TO REMEMBER

The same applies to all these type

of equations

4a2 – 36

= (2a -6)(2a+6)

Page 14: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

SOLVING USING THE QUADRATIC FORMULA

ax2 +bx +c is the standard form of a quadratic

equation (where a, b and c represent numbers)

to find x use the equation

x = (-b ± √(b2 – 4ac))/2a

Page 15: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

2x2 +8x + 6 = 0a = 2, b = 8 , c = 6

x = (-8 ± √(82 – 4*2*6))/2*2=(-8 ± √(64 – 48))/4

= ( -8 ± √16)/4(-8 ± 4)/4

= -12/4 or -4/4= -3 or -1

SOLVING USING THE QUADRATIC FORMULA

Page 16: QUADRATICS ax 2 + bx + c. MULTIPLYING BRACKETS (FOIL) x 2 +5x +6  Outside  ( x+ 3)(x + 2) First x2x2 Inside Last +2x+3x +6

SOLVING USING THE QUADRATIC FORMULA

x2 -10x + 34 = 0x = (10 ± √(102 – 4*1*-34))/2

=(10 ± √(100 – 136))/2= ( 10 ± √-36)/2

(10 ± 6i)/2= 5 + 3i or 5 – 3i

i is an imaginary number (i2 = -1)

Cannot havea zero square number so has to be multipliedby i2 to make it positive