Algebra: quadratic equations. Solve for \(x\): \(x^{2} - 5x - 6 = 0\)

1. \(x = 6\) or \(x = -1\)
2. \(x = -6\) or \(x = 1\)
3. \(x = 2\) or \(x = 3\)
4. \(x = 6\) or \(x = 1\)

Answer: \(x = 6\) or \(x = -1\)

Factorise: two numbers with product \(-6\) and sum \(-5\) are \(-6\) and \(+1\).

\((x - 6)(x + 1) = 0\)

Tip: Check the signs: the roots must multiply to \(-6\) and add to \(-5\).

Algebra: quadratic formula. Solve for \(x\), correct to TWO decimal places: \(2x^{2} - 4x - 3 = 0\)

1. \(x = 2{,}58\) or \(x = -0{,}58\)
2. \(x = 2{,}58\) or \(x = 0{,}58\)
3. \(x = 1{,}58\) or \(x = -0{,}58\)
4. \(x = 3{,}58\) or \(x = -1{,}58\)

Answer: \(x = 2{,}58\) or \(x = -0{,}58\)

\(a = 2,\ b = -4,\ c = -3\). Use \(x = \dfrac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\).

\(x = \dfrac{4 \pm \sqrt{16 + 24}}{4} = \dfrac{4 \pm \sqrt{40}}{4}\)

\(\sqrt{40} \approx 6{,}32\), so \(x = \dfrac{4 + 6{,}32}{4}\) or \(x = \dfrac{4 - 6{,}32}{4}\).

Tip: Use the formula whenever the trinomial will not factorise neatly.

Algebra: quadratic inequalities. Solve for \(x\): \(x^{2} - 2x - 15 \leq 0\)

1. \(-3 \leq x \leq 5\)
2. \(x \leq -3\) or \(x \geq 5\)
3. \(-5 \leq x \leq 3\)
4. \(x \leq -5\) or \(x \geq 3\)

Answer: \(-3 \leq x \leq 5\)

Factorise: \((x - 5)(x + 3) \leq 0\), so the critical values are \(x = 5\) and \(x = -3\).

The parabola opens upward, so it is below (or on) the \(x\)-axis BETWEEN the roots.

Tip: \(\leq 0\) means between the roots; \(\geq 0\) means outside them. A quick sketch settles it.

Algebra: surd equations. Solve for \(x\): \(\sqrt{x + 7} = x + 1\)

1. \(x = 2\)
2. \(x = 2\) or \(x = -3\)
3. \(x = -3\)
4. \(x = 3\)

Answer: \(x = 2\)

Square both sides: \(x + 7 = (x + 1)^{2} = x^{2} + 2x + 1\).

\(0 = x^{2} + x - 6 = (x + 3)(x - 2)\), so \(x = -3\) or \(x = 2\).

Test both: \(x = 2\) gives \(\sqrt{9} = 3 = 2 + 1\) (valid); \(x = -3\) gives \(\sqrt{4} = 2 \neq -2\) (rejected).

Tip: Always test in the ORIGINAL equation. Squaring can create a false solution.

Algebra: simplifying surds. Simplify: \(\sqrt{48} - \sqrt{12}\)

1. \(2\sqrt{3}\)
2. \(6\sqrt{3}\)
3. \(2\sqrt{6}\)
4. \(6\)

Answer: \(2\sqrt{3}\)

\(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}\).

\(\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}\).

Tip: You cannot subtract under the root: \(\sqrt{48} - \sqrt{12} \neq \sqrt{36}\).