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I was given this problem to solve with elementary methods (High School level).

Knowing that $$\begin{align} a^2+a b+b^2 &=40 \\ a^2-\sqrt{a b}+b &=\phantom{0}5 \end{align}$$ find $$a^2+\sqrt{a b}+b$$

I tried to look for $\sqrt{a b}$, since the requested quantity is

$$a^2+\sqrt{a b}+b=(a^2-\sqrt{a b}+b)+2\sqrt{ab}=5 + 2\sqrt{ab}$$ So I set $$x=a^2;\;y=\sqrt{ab};\;b=z$$ and the system became $$x+y^2+z^2=40;\;x-y-z=5$$ subtracting the two equations I got $$z^2-z+y^2+y-35=0$$ which has one real solution when the discriminant is zero.

That is

$1-4(y^2+y-35)=0$

and then

$y=\frac{1}{2} \left(-1\pm\sqrt{142}\right)$

and finally $$a^2+\sqrt{a b}+b=4\pm\sqrt{142}$$

I know that there are other solutions because I've found them with Wolfram Mathematica, but I couldn't find them with elementary methods.

Any help will be appreciated.

NadAlaba
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Raffaele
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    Very interesting your question and my regards from Sicily to the colleague. +1 – Sebastiano Aug 09 '20 at 16:00
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    Are you it is not $$a-\sqrt{ab}+b=5?$$ Then we can use $$(x^2-xy+y^2)(x^2+xy+y^2)=?$$ – lab bhattacharjee Aug 09 '20 at 16:06
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    @labbhattacharjee No, it is not. They sent me the photo of the book... – Raffaele Aug 09 '20 at 16:13
  • @Raffaele as you proceed, $1-4(y^2+y-35)= k \ge 0 $ then you can get solutions varry over $k\in[0,\infty) $ – A learner Aug 09 '20 at 16:16
  • As you can't get $1-4(y^2+y-35) \lt 0 $ , so, I think , over $k$ you will get all the possible solutions. – A learner Aug 09 '20 at 16:19
  • @Raffaele Are you only looking for real solutions? – Joe Aug 09 '20 at 19:33
  • Wolfram gives three possible values for $a$ which are roots of the polynomial $a^8-2a^7-16a^6+24a^5+33a^4+120a^3+240a^2-950a+225$. Moreover if $x=a^2+\sqrt{ab}+b$, then $(x-a^2+a)(5-a^2+a)=40$. The values of $x$ are roots of $x^{8} - 36 x^{7} + 604 x^{6} - 6620 x^{5} + 46638 x^{4} - 202780 x^{3} + 761740 x^{2} - 1350500 x - 7124375$. – Fabio Lucchini Aug 10 '20 at 15:20
  • You wrote "and finally $a^2+\sqrt{ab}+b=4\pm\sqrt{142}$. I know that there are other solutions because I've found them with Wolfram Mathematica". If you mean that Wolfram Mathematica showed that it is possible that $a^2+\sqrt{ab}+b=4\pm\sqrt{142}$, then what are $(a,b)$ such that $a^2+ab+b^2=40,a^2-\sqrt{ab}+b=5$ and $a^2+\sqrt{ab}+b=4\pm\sqrt{142}$ ? Also, what are "other solutions" ? I've got the same equation as the one written at the end of Fabio Lucchini's comment which implies that it is impossible that $a^2+\sqrt{ab}+b=4\pm\sqrt{142}$. – mathlove Aug 12 '20 at 05:51
  • @mathlove: The values $a^2+\sqrt{ab}+b=4\pm\sqrt{142}$ are wrong by comparison with approximate values given by wolfram. – Fabio Lucchini Aug 12 '20 at 07:59
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    Even with the photo of the book in hand, given what's been shown so far I think it's clear that Occam's Razor suggests the correct answer is 'the book has a typo'. The methods being thrown at this problem are far from high-school level. – Steven Stadnicki Sep 06 '20 at 20:10
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    I agree with the last comment. It is very likely that there is a typo in the book. Probably the correct version of the problem should have $a-\sqrt{ab}+b$ as the second equation and should ask for the value of $a+\sqrt{ab}+b$, which in this case would be $8$, as confirmed here: https://www.wolframalpha.com/input/?i=a%5E2%2Ba+b%2Bb%5E2+%3D40%2C+a-%5Csqrt%7Ba+b%7D%2Bb%3D5%2C+a%2B%5Csqrt%7Ba+b%7D%2Bb%3Dx+++++++ – Anatoly Sep 06 '20 at 20:43

2 Answers2

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We could try to "brute force" it at the high school level. Start with the second equation $$a^2 - \sqrt{ab} + b = 5$$ $$a^2 + b -5 = \sqrt{ab}$$ $$a^4 + b^2 + 25 + 2a^2 b -10a^2 -10b = ab$$ $$b^2 + ( 2a^2 -10 -a) b + (a^4 -10a^2 +25) =0$$ $$b = \frac{-(2a^2 -10 -a) \pm \sqrt{( 2a^2 -10 -a)^2 - 4(a^4 -10a^2 +25))}}{2}$$ $$b = \frac{-(2a^2 -10 -a) \pm \sqrt{a (-4 a^2 + a + 20)}}{2}$$ Now let's just call $b=x \pm \sqrt{y}.$ $$b^2 = x^2 \pm 2x\sqrt{y} + y$$ $$ab = ax \pm a\sqrt{y}$$ Plug this into the second equation: $$a^2 +ab + b^2 =40$$ $$a^2 + ax \pm a\sqrt{y} + x^2 \pm 2x\sqrt{y} + y =40 $$ $$(\pm a \pm 2x)\sqrt{y}=40 -a^2 - ax - x^2 -y$$ $$(a^2 \pm 2xa + 4x^2)y=(40 -a^2 - ax - x^2 -y)^2$$ Now plug in for $x, y$ and solve for $a$. What you end up with is: $$-225 + 950 a - 290 a^2 - (255 a^3)/2 - (53 a^4)/4 - (45 a^5)/2 + 14 a^6 + 2 a^7 - a^8 =0$$ This is not tractable, and it is likely the problem has a typo.

If you type in Mathematica

Solve[{a^2 + a*b + b^2 == 40, \[Sqrt](a*b) + 5 == a^2 + b}, {a, b}]

It will spit out 3 gross looking expressions possible answers. If you then evaluate these for $a^2 + b + \sqrt{ab}$

a^2 + b + Sqrt[a b] /. %OUTPUT NUMBER

Then the resulting expression doesn't simplify nicely. (Which is not a guarantee; Mathematica's simplify function is far from perfect).

Jbag1212
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  • I discovered (too late) that the problem, a problem printed on a BOOK, had a typo!!! Second equation was $$a^2+\sqrt{a b}+b^2$$. And the quantity to find was $$a^2-\sqrt{a b}+b^2$$. Which is almost trivial to solve... – Raffaele Nov 16 '20 at 10:57
  • Please add this comment to the question itself – sato Sep 30 '21 at 04:34
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One way to do this is to graph out these equations of lines. Find the values of $a,b$ and plug into the desired output equation:

  1. Equation of a line: see link here.

  2. Look at the possible values and

  3. plug answers into the final equation:

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Siong Thye Goh
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Amit Shah
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