MHB What is the difference between b and a in the given expression?

[Albert1]

$a,b\in R$

$if :\,\,5a^2+8ab+5b^2+170=50a+58b$

please find :$b-a$

 

[mente oscura]

Re: find b-a

$a,b\in R$

$if :\,\,5a^2+8ab+5b^2+170=50a+58b$

please find :$b-a$

Hello.

Spoiler

5a^2-a(50-8b)+5b^2-58b+170=0

a=\dfrac{50-8b \pm \sqrt{-36b^2-360b-900}}{10}

b=-5

\forall{b}>-5 \ and\ \forall{b}<-5 \rightarrow{b \cancel{\in{R}}}

If \ b=-5 \rightarrow{a \cancel{\in{R}}}

Conclusion:

\cancel{\exists}{a,b} \in{R} \ / \ 5a^2+8ab+5b^2+170=50a+58b

Regards.

 

[mathbalarka]

Re: find b-a

Untrue. A doable solution is :

Spoiler

(1, 5)

 

[mathbalarka]

Re: find b-a

I don't usually post solutions to elementary number theory, but doing so to point out mente oscura's flaw :

Spoiler

Going in the line of mente oscura, we have :

$$5a^2-a(50-8b)+5b^2-58b+170=0$$

which has the discriminant of $-36b^2+360b-900 = 36(5-b)^2$

This easily gives $b = 5$

 

[mente oscura]

Re: find b-a

Untrue. A doable solution is :

Spoiler

(1, 5)

Correct. Brute mistake. (Headbang)

Regards.

 

[Albert1]

Re: find b-a

$a,b\in R$

$if :\,\,5a^2+8ab+5b^2+170=50a+58b$

please find :$b-a$

solution:

Spoiler

$(2a+b)^2+(2b+a)^2+170=50a+58b---(1)$
let :$x=2a+b,\,\, y=(2b+a)$
then :$a=\dfrac{2x-y}{3},\,\, b=\dfrac{2y-x}{3}$
(1)becomes:$3(x-7)^2+3(y-11)^2=0$
we have :$x=7,\,\, y=11$
$\therefore y-x=b-a=4$