Question

Please help with this question and give detailed explanation!! thank you!!

Answer 1

`(a+b)/(a-b)=-√(2)`

Explanation:

Given:

`a^2+b^2=6ab`

`0 < a < b`

Add 2ab to both sides of the given equation:

`\implies a^2+b^2+2ab=6ab+2ab`

`\implies a^2+2ab+b^2=8ab`

Factor the left side:

`\implies (a+b)^2=8ab`

Subtract 2ab from both sides of the given equation:

`\implies a^2+b^2-2ab=6ab-2ab`

`\implies a^2-2ab+b^2=4ab`

Factor the left side:

`\implies (a-b)^2=4ab`

Therefore:

`\implies ((a+b)^2)/((a-b)^2)=(8ab)/(4ab)`

`\implies ((a+b)^2)/((a-b)^2)=2`

`\textsf{Apply exponent rule} \quad (a^c)/(b^c)=\left((a)/(b)\right)^c:`

`\implies \left((a+b)/(a-b)\right)^2=2`

Square root both sides:

`\implies \sqrt{\left((a+b)/(a-b)\right)^2}=√(2)`

`\implies (a+b)/(a-b)=\pm√(2)`

As 0 < a < b then:

• a + b > 0
• a - b < 0

Therefore:

`\implies (a+b)/(a-b)=(+)/(-)=-`

So:

`\implies (a+b)/(a-b)=-√(2)`

Answer 2

• 
  `-√(2)`

Explanation:

Make the following operations:

• a² + b² = 6ab
• a² + 2ab + b² = 8ab
• (a + b)² = 8ab
• a + b =
  `2√(2ab)`

and

• a² + b² = 6ab
• a² - 2ab + b² = 4ab
• (a - b)² = 4ab
• a - b =
  `-2√(ab)`, since a < b

The required value is:

• 
  `\cfrac{a+b}{a-b} =\cfrac{2√(2ab) }{-2√(ab) } =-√(2)`