Distinct Reals

Note that $\frac{a}{b}+\frac{b}{a}$ $=\frac{a^2}{ab}+\frac{b^2}{ab}$ $=\frac{a^2+b^2}{ab}$ $=\frac{(a+b)^2-2ab}{ab}.$ For a quadratic equation $Ax^2+Bx+C=0,$ where $A$ , $B$ and $C$ are real numbers and $A\neq 0,$ Vieta's Theorem states that $\text{Sum of roots} = -\frac{B}{A}$ and $\text{Product of roots} = \frac{C}{A}.$ It thus follows that $a+b=-\frac{3}{1}=-3$ and $ab=\frac{1}{1}=1.$ Hence, $\frac{(a+b)^2-2ab}{ab}$ $=\frac{(-3)^2-2(1)}{1}$ $=9-2=\boxed{7},$ and we are done. $_{\square}$

Given $\displaystyle { a }^{ 2 }+3a+1=0$ & $\displaystyle { b }^{ 2 }+3b+1=0$ . Clearly a and b are roots of the equation $\displaystyle { x }^{ 2 }+3x+1=0$ .

$\therefore a+b=-3\quad \& \quad ab=1$ We need to find $\cfrac { a }{ b } +\cfrac { b }{ a } =\cfrac { { a }^{ 2 }+{ b }^{ 2 } }{ ab } =\cfrac { { \left( a+b \right) }^{ 2 }-2ab }{ ab }$

Substituting the values we get $\cfrac { { \left( a+b \right) }^{ 2 }-2ab }{ ab } =\cfrac { { \left( -3 \right) }^{ 2 }-2 }{ 1 } =\cfrac { 7 }{ 1 }$ Hence the answer.

Since the equations given are the same one but with different variables we can deduce that $a$ and $b$ are the two roots of the equation $x^{2}+3x+1=0$ , which are $x_{+}=\frac{-3+\sqrt{5}}{2}$ $x_{-}=\frac{-3-\sqrt{5}}{2}$

Now we take $x_{+}=a$ and $x_{-}=b$ and begin to operate this way:

$\frac{a}{b}+\frac{b}{a}=$ $=\frac{\frac{-3+\sqrt{5}}{2}}{\frac{-3-\sqrt{5}}{2}}+\frac{\frac{-3-\sqrt{5}}{2}}{\frac{-3+\sqrt{5}}{2}}=$ $=\frac{-3+\sqrt{5}}{-3-\sqrt{5}}+\frac{-3-\sqrt{5}}{-3+\sqrt{5}}=$ $=\frac{(-3+\sqrt{5})(-3+\sqrt{5})+(-3-\sqrt{5})(-3-\sqrt{5})}{(-3-\sqrt{5})(-3+\sqrt{5})}=$ $=\frac{9+5-6\sqrt{5}+9+5+6\sqrt{5}}{9-5}=\frac{28}{4}=\boxed{7}$

x^2+3x+1=0 a+b=-3 ab=1 (a/b)+(b/a)=(a^2+b^2)/ab={(a+b)^2-2ab}/ab=9-2=7

a^2+3a+1=b^2+3b+1 a^2-b^2=-3(a-b) a+b=-3 ((a+b)^2-2ab)÷ab ((-3)^2-2(1))÷1 =7 Take note ab=1 since it is common on both the side with the a's and also at the b's.