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

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The discussion revolves around solving the equation 5a² + 8ab + 5b² + 170 = 50a + 58b to find the value of b - a. Participants acknowledge the complexity of the problem, with some pointing out errors in previous attempts at solutions. A correct approach is mentioned, emphasizing the need for careful calculations. The conversation highlights the importance of clarity in mathematical reasoning. Ultimately, the goal remains to determine the difference between b and a accurately.
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$a,b\in R$

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

please find :$b-a$
 
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Re: find b-a

Albert said:
$a,b\in R$

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

please find :$b-a$

Hello.

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.
 
Re: find b-a

Untrue. A doable solution is :

(1, 5)
 
Re: find b-a

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

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 $b = 5$
 
Re: find b-a

mathbalarka said:
Untrue. A doable solution is :

(1, 5)

Correct. Brute mistake. (Headbang)

Regards.
 
Re: find b-a

Albert said:
$a,b\in R$

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

please find :$b-a$
solution:
$(2a+b)^2+(2b+a)^2+170=50a+58b---(1)$
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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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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