Question

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Answer 1

Question

If a = (√5+√2)/(√5-√2) & b = (√5-√2)/(√5+√2) then prove that 3a²+4ab-3b² = 4+(56)/(3) x √10

ANSWER

Given : -

a = (√5+√2)/(√5-√2) & b = (√5-√2)/(√5+√2)

Required to prove : -

• 3a²+4ab-3b² = 4+(56)/(3)√10

Proof : -

a = (√5+√2)/(√5-√2) & b = (√5-√2)/(√5+√2)

We need to prove that;

• 3a²+4ab-3b² = 4+(56)/(3) x √10

Consider the LHS part ;

3a²+4ab-3b²

since,

• a = (√5+√2)/(√5-√2)
• b = (√5-√2)/(√5+√2)

This implies;

Before substituting these values let's rationalize the denominators of the values of a & b .

a = (√5+√2)/(√5-√2)

Multiply the numerator & denominator with √5+√2

(√5+√2)/(√5-√2) x (√5+√2)/(√5+√2)

[(√5+√2)²]/[(√5)²-(√2)²]

[(√5)²+(√2)²+2(√5)(√2)]/[5-2]

[5+2+2√10]/[3]

[7+2√10]/[3]

b = (√5-√2)/(√5+√2)

Multiply the numerator & denominator with √5-√2

(√5-√2)/(√5+√2) x (√5-√2)/(√5-√2)

[(√5-√2)²]/[(√5)²-(√2)²]

[(√5)²+(√2)²-2(√5)(√2)]/[5-2]

[5+2-2√10]/[3]

[7-2√10]/[3]

Hence,

• value of a = [7+2√10]/[3]
• value of b = [7-2√10]/[3]

Substituting these values in the LHS part;

3a²+4ab-3b²

3 x ([7+2√10]/[3])²+4([7+2√10]/[3])([7-2√10]/[3])-3 x ([7-2√10]/[3])²

3 x ([7+2√10]²/[3]²)+4([7+2√10][7-2√10])/[9])-3 x ([7-2√10]²/[3]²)

3 x ([7]²+[2√10]²+2[7][2√10])/([9])+4([7]²-[2√10]²)/(9)-3 x ([[7]²+[2√10]²-2[7][2√10])/[(9)]

3 x (49+40+28√10)/(9)+4(49-40)/(9)-3 x (49+40-28√10)/(9)

3 x (89+28√10)/(9)+4(9)/(9)-3 x (89-28√10)/(9)

(89+28√10)/(3)+4-(89-28√10)/(3)

(89+28√10)/(3)+4 (- 89+28√10)/(3)

4+(89+28√10)/(3)+(-89+28√10)/(3)

4+[(89+28√10)+(-89+28√10)]/[3]

4+[(89+28√10-89+28√10)]/[3]

4+[28√10+28√10]/[3]

4+[56√10]/[3]

4+(56)/(3) x √10

4+(56)/(3)√10

LHS = RHS

Hence Proved ! ✓

Answer 2

Answer:

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