Answer:
[tex]\dfrac{a+b}{a-b}=-\sqrt{2}[/tex]
Step-by-step explanation:
Given:
[tex]a^2+b^2=6ab[/tex]
[tex]0 < a < b[/tex]
Add 2ab to both sides of the given equation:
[tex]\implies a^2+b^2+2ab=6ab+2ab[/tex]
[tex]\implies a^2+2ab+b^2=8ab[/tex]
Factor the left side:
[tex]\implies (a+b)^2=8ab[/tex]
Subtract 2ab from both sides of the given equation:
[tex]\implies a^2+b^2-2ab=6ab-2ab[/tex]
[tex]\implies a^2-2ab+b^2=4ab[/tex]
Factor the left side:
[tex]\implies (a-b)^2=4ab[/tex]
Therefore:
[tex]\implies \dfrac{(a+b)^2}{(a-b)^2}=\dfrac{8ab}{4ab}[/tex]
[tex]\implies \dfrac{(a+b)^2}{(a-b)^2}=2[/tex]
[tex]\textsf{Apply exponent rule} \quad \dfrac{a^c}{b^c}=\left(\dfrac{a}{b}\right)^c:[/tex]
[tex]\implies \left(\dfrac{a+b}{a-b}\right)^2=2[/tex]
Square root both sides:
[tex]\implies \sqrt{\left(\dfrac{a+b}{a-b}\right)^2}=\sqrt{2}[/tex]
[tex]\implies \dfrac{a+b}{a-b}=\pm\sqrt{2}[/tex]
As 0 < a < b then:
Therefore:
[tex]\implies \dfrac{a+b}{a-b}=\dfrac{+}{-}=-[/tex]
So:
[tex]\implies \dfrac{a+b}{a-b}=-\sqrt{2}[/tex]
