Answer:
[tex]\sf \dfrac{a+b}{a-b}= -\sqrt{2}[/tex]
Step-by-step explanation:
[tex]\sf a^2 + b^2 = 6ab \qquad where\:\:0 < a < b[/tex]
Basic formula:
[tex]\sf (a + b)^2 = a^2 +b^2 +2ab[/tex]
[tex]\sf (a - b)^2 = a^2 + b^2 -2ab[/tex]
solve for a + b
[tex]\sf \rightarrow a^2 + b^2 = 6ab[/tex]
[tex]\sf \rightarrow (a+b)^2-2ab = 6ab[/tex]
[tex]\sf \rightarrow (a+b)^2 = 6ab+2ab[/tex]
[tex]\sf \rightarrow (a+b)^2 = 8ab[/tex]
[tex]\sf \rightarrow a+b = \pm \sqrt{8ab}[/tex]
solve for a - b
[tex]\sf \rightarrow a^2 + b^2 = 6ab[/tex]
[tex]\sf \rightarrow (a-b)^2 + 2ab = 6ab[/tex]
[tex]\sf \rightarrow (a-b)^2 = 6ab-2ab[/tex]
[tex]\sf \rightarrow (a-b)^2 = 4ab[/tex]
[tex]\sf \rightarrow a-b = \pm \sqrt{4ab}[/tex]
Now solve for (a + b)/(a - b)
[tex]\sf \rightarrow \dfrac{a+b}{a-b}= \dfrac{\pm\sqrt{8ab} }{\pm\sqrt{4ab} }[/tex]
[tex]\sf \rightarrow \dfrac{a+b}{a-b}= \pm \sqrt\dfrac{8ab }{4ab}[/tex]
[tex]\sf \rightarrow \dfrac{a+b}{a-b}= \pm \sqrt{2}[/tex]
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.
[tex]\sf \rightarrow \dfrac{a+b}{a-b}= - \sqrt{2}[/tex]
