Revision (Grade 11 NSC Matric Mathematics): Revision Notes

Worked Example 1: Basic factorisation

Question: Solve $x(x - 3) = 10$

Solution:

First, expand and rearrange into standard form:

• $x^2 - 3x = 10$
• $x^2 - 3x - 10 = 0$

Next, factorise the quadratic expression. We need two numbers that multiply to give -10 and add to give -3. These numbers are -5 and +2.

• $(x + 2)(x - 5) = 0$

Apply the zero product law:

• $x + 2 = 0$ or $x - 5 = 0$
• $x = -2$ or $x = 5$

Therefore $x = -2$ or $x = 5$.

Worked Example 2: Standard form factorisation

Question: Solve $2x^2 - 5x - 12 = 0$

Solution:

The equation is already in standard form with no common factors.

To factorise, we need factors of $2 \times (-12) = -24$ that add to give -5. These are -8 and +3.

We can write: $2x^2 - 8x + 3x - 12 = 0$

Grouping: $2x(x - 4) + 3(x - 4) = 0$

Factoring: $(2x + 3)(x - 4) = 0$

Applying the zero product law:

$2x + 3 = 0$ or $x - 4 = 0$

$x = -\frac{3}{2}$ or $x = 4$

Therefore $x = -\frac{3}{2}$ or $x = 4$.

Worked Example 3: Difference of squares

Question: Solve $y^2 - 7 = 0$

Solution:

Recognise this as a difference of squares pattern where $(\sqrt{7})^2 = 7$.

Rewrite as: $y^2 - (\sqrt{7})^2 = 0$

Factorise using the difference of squares formula:

$(y - \sqrt{7})(y + \sqrt{7}) = 0$

Apply the zero product law:

$y - \sqrt{7} = 0$ or $y + \sqrt{7} = 0$

$y = \sqrt{7}$ or $y = -\sqrt{7}$

Therefore $y = \pm\sqrt{7}$.

Worked Example 4: Rational equations

Question: Solve $\frac{3b}{b + 2} + 1 = \frac{4}{b + 1}$

Solution:

First, identify restrictions. The denominators cannot equal zero, so $b ≠ -2$ and $b ≠ -1$.

Find the lowest common denominator: $(b + 2)(b + 1)$

Multiply each term by the LCD and simplify:

$\frac{3b(b + 1)}{(b + 2)(b + 1)} + \frac{(b + 2)(b + 1)}{(b + 2)(b + 1)} = \frac{4(b + 2)}{(b + 1)(b + 2)}$

This gives us: $3b(b + 1) + (b + 2)(b + 1) = 4(b + 2)$

Expanding: $3b^2 + 3b + b^2 + 3b + 2 = 4b + 8$

Simplifying: $4b^2 + 6b + 2 = 4b + 8$

Rearranging: $4b^2 + 2b - 6 = 0$

Dividing by 2: $2b^2 + b - 3 = 0$

Factorising: $(2b + 3)(b - 1) = 0$

Solving: $2b + 3 = 0$ or $b - 1 = 0$

$b = -\frac{3}{2}$ or $b = 1$

Both solutions satisfy our restrictions, so $b = -1\frac{1}{2}$ or $b = 1$.

Worked Example 5: Equations involving surds

Question: Solve $m + 2 = \sqrt{7 + 2m}$

Solution:

When dealing with equations containing square roots, we must isolate the radical on one side, then square both sides carefully.

Square both sides: $(m + 2)^2 = (\sqrt{7 + 2m})^2$

Expanding: $m^2 + 4m + 4 = 7 + 2m$

Rearranging: $m^2 + 2m - 3 = 0$

Factorising: $(m - 1)(m + 3) = 0$

So $m = 1$ or $m = -3$

Important: When squaring both sides, we may introduce invalid solutions, so we must check both answers.

For $m = 1$:

LHS = $1 + 2 = 3$

RHS = $\sqrt{7 + 2(1)} = \sqrt{9} = 3$ ✓

For $m = -3$:

LHS = $-3 + 2 = -1$

RHS = $\sqrt{7 + 2(-3)} = \sqrt{1} = 1$ ✗

Therefore $m = 1$ is the only valid solution.