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#### Albert

##### Well-known member

- Jan 25, 2013

- 1,225

$a,b\in R$

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

please find :$b-a$

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

please find :$b-a$

- Thread starter Albert
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- Thread starter
- #1

- Jan 25, 2013

- 1,225

$a,b\in R$

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

please find :$b-a$

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

please find :$b-a$

- Nov 29, 2013

- 172

Hello.$a,b\in R$

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

please find :$b-a$

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

[tex]b=-5[/tex]

[tex]\forall{b}>-5 \ and\ \forall{b}<-5 \rightarrow{b \cancel{\in{R}}}[/tex]

[tex]If \ b=-5 \rightarrow{a \cancel{\in{R}}}[/tex]

Conclusion:

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

Regards.

- Mar 22, 2013

- 573

Untrue. A doable solution is :

(1, 5)

- Mar 22, 2013

- 573

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

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

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

This easily gives $b = 5$

- Nov 29, 2013

- 172

Correct. Brute mistake.Untrue. A doable solution is :

(1, 5)

Regards.

- Thread starter
- #6

- Jan 25, 2013

- 1,225

solution:$a,b\in R$

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

please find :$b-a$

$(2a+b)^2+(2b+a)^2+170=50a+58b---(1)$

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 :$x=7,\,\, y=11$

$\therefore y-x=b-a=4$

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