Quadratic equations

Grade 11 – Paper 1

Quadratic equations

• Quadratic equations in one variable are equations of degree 2, which means the highest power of the variable is 2.

• The standard form is

ax2 + bx + c = 0 where a ≠ 0

• These equations have 2 solutions,

they can be:– different– the same– non-real

• The solutions are known as the roots of the equation.

Method:

• The solution of quadratic equations by factorising is based on the fact that if

p.q = 0 then either p or q must be zero.

Therefore there will be 2 solutions.

*This only works if the RHS is 0

Solve for x:

1. x – 3 = 02. (x – 3)(x + 2) = 03. x(x + 7) = 04. x2 - 4x = 05. x2 + x – 6 = 06. (x – 3)(x +5)(x – 1) = 0

If ONE factor is equal to zero, the whole expression will be zero.

Method 1: Factorising

1. x2 – 2x – 35 = 0 (x – 7)(x + 5) = 0

x – 7 = 0 or x = 7

2) x(x – 1) = 20x2 – x – 20 = 0

(x – 5)(x + 4) = 0

∴ x – 5 = 0 x = 5

x + 5 = 0 x = -5 x + 4 = 0

x = -4

Example 3

(x + 1)(x – 2) +3(x – 1)(x + 1) = 3(x – 2)

x2 – 2x + x – 2 + 3x2 – 3 = 3x – 6 4x2 – 4x + 1 = 0 (2x – 1)(2x – 1) = 0

2x – 1 = 0

2x = 1

x = ½OR

2x – 1 = 0

2x = 1

x = ½

Both roots are equal

Pg 68 Ex 4.29, 11, 17, 21, 24

Classwork

Ex 4.2 (9 & 11)

9) 3x2 – 12 = 0 3(x2 – 4) = 0 3(x – 2)(x + 2) =

0

x – 2 = 0 or x = 2

11) 2x2 = 18 (÷2) x2 = 9

x = ±√9 ∴ x = ±3x + 2 = 0

x = -2

Remember a square root have a

positive and a negative answer.

Ex 4.2 (17 & 21)

17) 7x - x2 – 6 = 0 - x2 + 7x – 6 =

0 -(x2 – 7x + 6) = 0 -(x – 6)(x – 1) = 0

x – 6 = 0 or x = 6

21) x(x – 1) = 4(3x – 10) x2 – x = 12x - 40

x2 – 13x + 40 = 0

(x – 5)(x – 8) = 0

∴ x – 5 = 0 or x – 8

= 0 x = 5 x

= 8

x – 1= 0 x = 1

Ex 4.2 (24)

24) 2(m – 1)(m +1)= 0 2m2 – 2 = 7m +14 -

1 2m2 – 7m – 15 = 0 (2m + 3)(m – 5) =

0

2m + 3= 0 or m – 5 = 0

2m = -3 m = 5

m = -3/2

Steps:1. Multiply out.2. Make the RHS = 03. Factorise LHS4. Set each factor =

05. Solve each factor.

Pg 68 Ex 4.22, 4, 7, 8,12, 13, 16, 18, 23

Home work

Equations with fractions

• Division by zero is undefined.– Start by writing down the

restrictions.

• Multiply with the denominator.

Example 1Limits:

x ≠ 0multiply with

x

(x – 5)(x + 2) = 0

(x – 5) = 0x = 5

(x + 2) = 0x = -2

Example 2

Limits:

x ≠ 2

multiply with (x-2)(x-3)

(x – 4) = 0x = 4

(x - 2) = 0x = 2

x ≠ 3(x – 4)(x – 2) = 0

Example 3

Limits:

x ≠ 2

multiply with (x+2)(x-2)(x+1)

(x – 3) = 0

x = 3

(x + 1) = 0

x = -1

x ≠ -2x ≠ -1

Pg 70 Ex 4.32, 4, 6, 10, 17

Home work

Method 3: Substitution(another k-method)

If an algebraic expression is repeated in the equation we can replace it with “k” to make a simpler equation.

Just remember to substitute back once you have calculated k.

1. (x2 + 2x)2 – 2(x2 + 2x) – 3 = 0

Let k = (x2 + 2x)

∴ k2 – 2k – 3 = 0 (k – 3)(k + 1) = 0

∴k = 3 or k = -1k = (x2 + 2x) 3 = x2 + 2x 0 = x2 + 2x - 30 = (x + 3)(x – 1)x = -3 or x = 1

k = (x2 + 2x) -1 = x2 + 2x 0 = x2 + 2x + 10 = (x + 1)(x + 1)x = -1 or x = -1

2. (x2 + x)2 – 14(x2 + x) + 24 = 0

Let k = (x2 + x)

∴ k2 – 14k + 24 = 0 (k – 12)(k – 2) = 0

∴k = 12 or k = 2k = (x2 + x) 12 = x2 + x 0 = x2 + x - 120 = (x + 4)(x – 3)x = -4 or x = 3

k = (x2 + x) 2= x2 + x 0 = x2 + x – 20 = (x - 1)(x + 2)x = 1 or x = -2

3.

Let k = (x2 - 3x)

k2 – 8k - 20 = 0 (k – 10)(k + 2) = 0

∴k = 10 or k = -2

k = (x2 - 3x) 10 = x2 – 3x 0 = x2 – 3x - 100 = (x + 2)(x – 5)x = -2 or x = 5

k = (x2 - 3x) -2 = x2 – 3x 0 = x2 – 3x + 2 0 = (x – 2)(x – 1)x = 2 or x = 1

Limits:

x ≠ 0x ≠ 3

Pg 72 Ex 4.42, 3, 7, 8, 9

Home work

Method 4: Squaring both sides

Squaring both sides might introduce a extra solution which is invalid.

x = 3 or x = -2

x = 3

x = -2

x ≠ -2

1) Test:

Notes:• By definition is non-negative.∴ √9 = 3 and √9 ≠ -3, but x2 = 9 √x2 = √9 x = ±3• The square root of a negative number is not defined.

√x ∈ ℝ ⇒ x ≥ 0• (a + b)2 ≠ a2 + b2 (a + b)2 = a2 + 2ab + b2

Limits:

x – 3 ≥ 0x ≥ 3

Pg 74 Ex 4.51, 4, 5, 7, 9, 10, 11, 12

Home work