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

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Question 1 (Pythagoras theorem) Step 1: Identify the right triangle PQR right-angled at Q, with PQ = 7 cm and QR = 5 cm. x = PR. By Pythagoras theorem: PR² = PQ² + QR² Step 2: Substitute the values. x² = 7² + 5²; x² = 49 + 25 Step 3: Simplify. x² = 74; x = √(74) √(74) cm (since length is positive). √(74) Question 3 (Tangents to a circle) Given: PA = PB = 10 cm (tangents from P), ∠ APB = 60°, O is centre. Find AO = r cm. Step 1: OA PA and OB PB (radius tangent). OP bisects ∠ APB, so ∠ OPA = ∠ OPB = 30°. Step 2: Consider right OAP (right-angled at A). tan(∠ OPA) = OA/PA; tan 30° = r/10; Step 3: Substitute tan 30° = (1)/(√(3)). (1)/(√(3)) = r/10 Step 4: Solve for r. r = (10)/(√(3)) Rationalize denominator: r = (10)/(√(3)) × (√(3))/(√(3)) = (10√(3))/(3) (10√(3))/(3) Question 5 (Quadratic equation) Solve x² − 5x + 6 = 0. Step 1: Factorize. Find two numbers that multiply to 6 and add to −5: −2 and −3. x² − 5x + 6 = (x − 2)(x − 3) = 0 Step 2: Solve. x − 2 = 0 or x − 3 = 0; x = 2 or x = 3 2, 3

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

Step-by-step solution

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

Substitute the values.

Simplify.

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

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

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

Solve for $r$ .

Factorize.

Solve.

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