Question

Please help with this question, thank you!!​

Answer 1

`-√(2)`

Explanation:

Note that:

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

Answer 2

`\sf (a+b)/(a-b)= -√(2)`

Explanation:

`\sf a^2 + b^2 = 6ab \qquad where\:\:0 < a < b`

Basic formula:

`\sf (a + b)^2 = a^2 +b^2 +2ab`

`\sf (a - b)^2 = a^2 + b^2 -2ab`

solve for a + b

`\sf \rightarrow a^2 + b^2 = 6ab`

`\sf \rightarrow (a+b)^2-2ab = 6ab`

`\sf \rightarrow (a+b)^2 = 6ab+2ab`

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

`\sf \rightarrow a+b = \pm √(8ab)`

solve for a - b

`\sf \rightarrow a^2 + b^2 = 6ab`

`\sf \rightarrow (a-b)^2 + 2ab = 6ab`

`\sf \rightarrow (a-b)^2 = 6ab-2ab`

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

`\sf \rightarrow a-b = \pm √(4ab)`

Now solve for (a + b)/(a - b)

`\sf \rightarrow (a+b)/(a-b)= (\pm√(8ab) )/(\pm√(4ab) )`

`\sf \rightarrow (a+b)/(a-b)= \pm \sqrt(8ab )/(4ab)`

`\sf \rightarrow (a+b)/(a-b)= \pm √(2)`

As, the value of b is greater than the value of a and both the values a and b are greater than 0. It turns the value of expression negative as the denominator will be always evaluated to negative integer and the numerator value always > 0. So, the answer will be negative, -√2.

`\sf \rightarrow (a+b)/(a-b)= - √(2)`