Question

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

Answer 1

Answer:

• $-\sqrt{2}$

Step-by-step explanation:

Make the following operations:

• a² + b² = 6ab
• a² + 2ab + b² = 8ab
• (a + b)² = 8ab
• a + b = $2\sqrt{2ab}$

and

• a² + b² = 6ab
• a² - 2ab + b² = 4ab
• (a - b)² = 4ab
• a - b = $-2\sqrt{ab}$, since a < b

The required value is:

• $\cfrac{a+b}{a-b} =\cfrac{2\sqrt{2ab} }{-2\sqrt{ab} } =-\sqrt{2}$

Answer 2

Answer:

$\dfrac{a+b}{a-b}=-\sqrt{2}$

Step-by-step 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:

$\implies \dfrac{(a+b)^2}{(a-b)^2}=\dfrac{8ab}{4ab}$

$\implies \dfrac{(a+b)^2}{(a-b)^2}=2$

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

$\implies \left(\dfrac{a+b}{a-b}\right)^2=2$

Square root both sides:

$\implies \sqrt{\left(\dfrac{a+b}{a-b}\right)^2}=\sqrt{2}$

$\implies \dfrac{a+b}{a-b}=\pm\sqrt{2}$

As 0 < a < b then:

• a + b > 0
• a - b < 0

Therefore:

$\implies \dfrac{a+b}{a-b}=\dfrac{+}{-}=-$

So:

$\implies \dfrac{a+b}{a-b}=-\sqrt{2}$