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Examples and notes on solving quadratic equations
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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