Quadratics – Solving more complex quadratic equations

• Factorising (use when possible)
• Completing the square
• Quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Solve: \[ \frac{x}{3} - \frac{1}{x} = 2 \]

Step 1: Multiply through by \(3x\):

\[ x^2 - 3 = 6x \Rightarrow x^2 - 6x - 3 = 0 \]

Step 2: Use the formula:

\[ x = \frac{6 \pm \sqrt{36 + 12}}{2} = \frac{6 \pm \sqrt{48}}{2} = \frac{6 \pm 4\sqrt{3}}{2} = 3 \pm 2\sqrt{3} \]

Solve: \[ 2x^2 + 8x - 5 = 0 \]

Step 1: Factor out the 2:

\[ 2(x^2 + 4x) - 5 = 0 \Rightarrow x^2 + 4x = \frac{5}{2} \]

Step 2: Complete the square:

\[ x^2 + 4x + 4 = \frac{5}{2} + 4 = \frac{13}{2} \Rightarrow (x + 2)^2 = \frac{13}{2} \]

Step 3: Square root both sides:

\[ x + 2 = \pm \sqrt{\frac{13}{2}} \Rightarrow x = -2 \pm \sqrt{\frac{13}{2}} \]

Find the values of \(k\) for which: \[ x^2 - 4x + k = 0 \] has real solutions.

Method: Use the discriminant \(b^2 - 4ac \ge 0\).

\[ b^2 - 4ac = (-4)^2 - 4(1)k = 16 - 4k \] Real solutions require: \[ 16 - 4k \ge 0 \Rightarrow k \le 4 \]

• If factorising looks messy → use the quadratic formula.
• A negative discriminant means no real solutions.
• Simplify square roots when possible (\(\sqrt{48} = 4\sqrt{3}\)).
• When completing the square, factor out the coefficient of \(x^2\) first.

1. \(\frac{3}{x} - \frac{2}{x^2} = 1\)
2. \(3x^2 - 7x + 4 = 0\)
3. \(x^2 + (k-2)x + 3 = 0\) —— find values of \(k\) for which there are no real roots.

Give full working and simplify roots.