How Real Are These Fractions?

Relevant wiki: Trivial Inequality

No quadratics is needed! From the equation,

$(a+b)^2=ab$

$a^2+2ab+b^2=ab$

$a^2+b^2=-ab$

If $ab=0$ , both $a$ and $b$ must be 0, but since $a+b$ cannot be zero, this cannot be true.

Since squares of real numbers are non-negative, from the first and third equation, both $ab$ and $-ab$ are zero or positive at the same time. We have proved that $ab\neq0$ , therefore a contradiction.

Relevant wiki: Quadratic Discriminant - Problem Solving

Note that $a, a+b \neq 0$ because they appear on the denominator. If $b = 0$ , then the equation reduces to $\frac{a}{a} = 0$ , contradiction. Thus $b \neq 0$ either. So we can cross-multiply to obtain $(a+b)^2 = ab$ , or $a^2 + ab + b^2 = 0$ . Dividing by $b^2$ gives $\left( \frac{a}{b} \right)^2 + \frac{a}{b} + 1 = 0$ . This has no solution of $\frac{a}{b}$ on the reals (the discriminant is $1^2 - 4 \cdot 1 \cdot 1 = -3$ ). But if $a,b$ are reals, then $\frac{a}{b}$ must also be a real number, contradiction. So there is no real solution for $a,b$ .

Relevant wiki: Using the Quadratic Formula

Starting off, multiply by $a(a+b)$ on both sides. This gives $(a+b)^2=ab$ . After simplifying and gathering like terms we have $a^2+ab+b^2=0$ . We can now use the quadratic formula, giving $a=\dfrac{-b\pm\sqrt{-3b^2}}{2}$ . Because there is a negative inside the radical, either $a$ or $b$ must be imaginary to make the equation​ correct. Therefore there are no real pairs of numbers $a$ and $b$ that are a solution to the given equation.