Factorising & solving quadratic equations · VCE Maths Methods - Unit 1 - Factorising & solving quadratic equations The quadratic formula • The solution to the general quadratic

VCE Maths Methods - Unit 1 - Factorising & solving quadratic equations

Factorising & solving quadratic equations

• Expanding quadratic expressions• Factorising quadratic expressions• Factorisation by inspection• Completing the square• Solving quadratic equations• The quadratic formula• The discriminant


Expanding quadratic expressions

Each term in one bracket must be multiplied by the terms in the other bracket.

y =(x−1)(x+5) y =2(3x−1)(x−4) y =(2x+5)2

y = x2+5x−x−5

y = x2+4x−5

y =2(3x2 −12x−x+4)

y =2(3x2 −13x+4)

y =6x2 −26x+8

y =4x2+10x+10x+25

y =4x2+20x+25

“FOIL” - !rst, outside, inside,

last

(ax+b )2 =(a2x2+2abx+b2 )

(ax−b )2 =(a2x2 −2abx+b2 )

y =(x+4)(x−4)

y = x2+4x−4x−16

y = x2 −16

This is a perfectsquare.

This is a di"erence of two squares.

(ax+b )(ax−b )=(a2x2 −b2 )


Factorising quadratic expressions

Find highest common factor

Perfect squares

Factorisation by inspection

(This is now much easier to factorise.)

y = x2+6x+9

This is the square of (x+3)

y =(x+3)2

9 =3 2×3=6

y =60x2+40x+5

y =5(12x2+8x+1) y =5(6x+1)(2x+1)

Di"erence of two squares

y =49x2 −4

y =72 x2 −22

y =(7x+2)(7x−2)

y =5(12x2+8x+1)


Factorisation by inspection

No common factors

y = x2+16x+63

y =16x2 −24x+9

y =6x−24x2

y =72x2 −24x+2

Take out the common factor of 6x

y =6x(1−4x ) y =(x+7)(x+9)

7 & 9 are the only factors of 63

y =16x2 −24x+9Factors of 16: 1 & 16, 2 & 8, 4 & 4

Factors of 9: 1 & 9, 3 & 3

y =(16x− .....)(x− .....) y =(4x− .....)(4x− .....)

y =(4x−1)(4x−9)

y =(4x−3)(4x−3)=(4x−3)2

There are lots of factor pairs from 36, but only one pair from 1.

(And they must both be negative!)

y =(.....x−1)(.....x−1)

y =2(6x−1)2

y =2(36x2 −12x+1)


Completing the square

• Any quadratic function can be made into a perfect square form.

• This can be useful to !nd turning points or to solve di#cult equations.

y =2x2+12x+14

y =2(x2+6x+7)

y =2(x2+6x+9−9+7)

y =2 (x+3)2 −2⎡⎣ ⎤⎦

y =2(x+3)2 −4

The co-e#cient of x2 must be 1, 2 is taken out as a common factor.

The co-e#cient of x is halved, squared,added & subtracted. (No change to equation)

y =2 (x2+6x+9)−9+7⎡⎣ ⎤⎦ The !rst three terms make a perfect square.

2 is multiplied back in.


Solving quadratic equations

• The quadratic equation needs to !rst be factorised.

• Quadratic equations are solved using the Null factor law - if either factor is equal to 0, then the whole equation is equal to 0.

• On the graph, the solutions to the equation y = 0 are the x intercepts.

0=(x−4)2 0=(x−2)(2x+5) 0= x2+6x−12

0= x−2x =2

0=2x+5−5=2x

x =−52

0= x−4x =4

This is a repeated factor & just one solution.

(The graph turns on the x axis, without crossing)

0= x2+6x+9−9−12

0=(x+3)2 −21

0=(x+3)2 −( 21)2

0=(x+3+ 21)(x+3− 21)

0= x+3+ 21

x =−3− 21

0= x+3− 21

x =−3+ 21

This can be a DOTS!


The quadratic formula

• The solution to the general quadratic formula (0 = ax2 + bx + c) can be found by completing the square.

• This can be used to !nd any solutions that exist for a given quadratic.

y = ax2+bx+c

x = −b± b2 −4ac

2a

a=2, b=-4, c=-6 y =2x2 −4x−6For example:

x =

−−4± (−4)2 −(4×2×−6)2×2

x = −b± b2 −4ac

2a

x = 4± 64

4 x = 4±8

4x = -1 & x = 3 (Two solutions)


The discriminant

• Quadratic functions can have two, one or no solutions.

• The number of solutions can be determined by the discriminant.

• This is the expression inside the square root in the quadratic equation.

• If, $ < 0, there is no solution.

• If, $ = 0, there is one solution.

• If, $ > 0, there are two solutions.

Δ=b2 −4ac


The discriminant

y = x2 −8x+16

Δ=(−8)2 −(4×1×16)=0

y = x2 −6x+11

Δ=(−6)2 −(4×1×11)=−8

y =(x−4)2

y =(x−3)2+2 y =(x−5)2 −4

y = x2 −10x+21

Δ=(−10)2 −(4×1×21)=16

One solution No solutions Two solutions


The discriminant

• For what values of k does the quadratic function y=x2-4kx+20 have one, two or no solutions?

• The number of solutions can be determined by the discriminant.

a=1,b =−4k ,c =20 Δ=b 2 −4ac

Δ= (−4k )2 −4×1×20=0

16k 2 −80=0

80=16k 2

5=k 2

k = ± 5

If, $ = 0, there is one solution:

If, $ < 0, there is no solution:

If, $ > 0, there are two solutions:

k > 5 andk <− 5

k2 <5

− 5 <k < 5

k2 >5

16k 2 −80>0

16k 2 −80<0


The discriminant

y = x 2 −(4×2)x +20

Δ= (4 5)2 −(4×1×20)=0

y = x 2 −(4 5)x +20

Δ= (4 5)2 −(4×1×20)=0

k > 5 k = 5 − 5 <k < 5

y = x 2 −(4×3)x +20

Δ=(−10)2 −(4×1×21)=16

No solutionsOne solution Two solutions

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Factorising & solving quadratic equations · VCE Maths Methods - Unit 1 - Factorising & solving quadratic equations The quadratic formula • The solution to the general quadratic