Solve the equation: 4x2 - 4(a2-b2)x + (a4-b4) = 0

Question 1 (Pythagoras theorem)

Step 1: Identify the right triangle $\triangle PQR$ right-angled at $Q$, with $PQ = 7$ cm and $QR = 5$ cm. $x = PR$.

By Pythagoras theorem:

$PR^{2} = PQ^{2} + QR^{2}$

Step 2: Substitute the values.

$x^{2} = 7^{2} + 5^{2}$

$x^{2} = 49 + 25$

Step 3: Simplify.

$x^{2} = 74$

$x = \sqrt{74}$

$\sqrt{74}$ cm (since length is positive).

\sqrt{74}

Question 3 (Tangents to a circle)

Given: $PA = PB = 10$ cm (tangents from $P$), $\angle APB = 60^\circ$, $O$ is centre. Find $AO = r$ cm.

Step 1: $OA \perp PA$ and $OB \perp PB$ (radius $\perp$ tangent).

$OP$ bisects $\angle APB$, so $\angle OPA = \angle OPB = 30^\circ$.

Step 2: Consider right $\triangle OAP$ (right-angled at $A$).

$\tan(\angle OPA) = \frac{OA}{PA}$

$\tan 30^\circ = \frac{r}{10}$

Step 3: Substitute $\tan 30^\circ = \frac{1}{\sqrt{3}}$.

$\frac{1}{\sqrt{3}} = \frac{r}{10}$

Step 4: Solve for $r$.

$r = \frac{10}{\sqrt{3}}$

Rationalize denominator:

$r = \frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{10\sqrt{3}}{3}$

\dfrac{10\sqrt{3}{3}}

Question 5 (Quadratic equation)

Solve $x^{2} - 5x + 6 = 0$.

Step 1: Factorize.

Find two numbers that multiply to $6$ and add to $-5$: $-2$ and $-3$.

$x^{2} - 5x + 6 = (x - 2)(x - 3) = 0$

Step 2: Solve.

$x - 2 = 0 \quad or \quad x - 3 = 0$

$x = 2 \quad or \quad x = 3$

2,\ 3