Question

a and b are real (+), and two equations of x^2 + ax + 2b = 0 and x^2 + 2bx + a = 0 has real roots. find the minimum value of a + b = ... ?

Answer 1

a and b must be > 0
how possible get a + b = - 8 ?

Answer 2

when a> 0 and b>0 , a+b=-8 is not possible at all

Answer 3

https://www.desmos.com/calculator/kkrxgkali2

Answer 4

from the graph minimum seems 0

Answer 5

for the first one ,it needs discriminant \(\geq 0\) to have real roots, that is \(a^2-8b \geq 0\) or \(a^2 \geq 8b\)
for the second one, exactly the same, we have \(b^2-a\geq 0\), that is \(b^4\geq a^2\)

Answer 6

Now combine: \(8b \leq a^2\leq b^4\), that gives \(8b\leq b^4\), hence
\(b^4-8b \geq 0\) , hence \(b(b-2)(b^2+2b+4)\geq 0\)
b>0 (given), hence, (b-2) and (b^2+2b+4) must be both positive or both negative

Answer 7

so, what is the answer ?

Answer 8

but b^2 +2b +4 is always positive, it means you don't have roots for this, in other word, you don't have value of b here, so that just b-2>0 --> b>2
plug back to \(a\leq b^2\) that is the maximum value of a is b^2 =4
hence \(a+b\leq 6\)