Prove that the equation has no real roots

Question

Given $a>0$ and $b>0$, how can we prove that $x^2 - (a+b)x +2ab=0$ has no real roots?

I thought about starting with the discriminant.

I calculated the discriminant and found $\Delta=(a+b)^2-8ab $.

How do I show that $\Delta$ is less than zero?

Answer 1

We need to prove that $$(a+b)^2-8ab<0,$$ which is wrong.

Take $b=0.1$ and $a=1$.

Answer 2

\begin{align} & (a+b)^2 - 8ab = (a^2 + 2ab + b^2) - 8ab \\[10pt] = {} & \underbrace{(a^2 -6ab) + b^2 = (a^2 -6ab + 9b^2) - 8b^2}_{\large\text{completing the square}} = (a-3b)^2 -8b^2 \end{align} If $a$ is large and $b$ is small then this is positive.

Answer 3

Multiply by $4$ to obtain $$4x^2-4(a+b)x+8ab=(2x-a-b)^2+8ab-(a+b)^2=(2x-a-b)^2=$$$$=(2x-a-b)^2+(a+b)^2-2(a-b)^2$$

The negative square on the right-hand side is a sign that the expression is not definite (ie will take both positive and negative values). For example, take $a=6b$ then it becomes $$(2x-8b)^2-b^2$$ which is negative when $x=4b$.

Answer 4

Well, why not. We want to investigate when $a^2 - 6ab+b^2$ is positive, giving counterexamples to the question. We can start with $(1,0),$ so that $a^2 - 6ab+b^2= 1.$

You might complain that $a,b$ are not both positive. That will be alright, given an $(a,b),$ we get a new pair with $$ (a,b) \mapsto (6a-b,a) \; . $$ The value of $a^2 - 6ab + b^2$ does not change when doing this. So, here are an infinite sequence of pairs with $a^2 - 6ab + b^2 = 1 \; :$ $$ (1,0), \; \; (6,1), \; \; (35,6), \; \; (204, 35),\; \; (1189, 204), \ldots $$ These also follow separate rules for the two letters, namely $$ a_{n+2} = 6 a_{n+1} - a_n, $$ $$ b_{n+2} = 6 b_{n+1} - b_n. $$