Wolfram tells me that the solution to \begin{equation*} \sqrt{42}\sqrt{(x-6)(x+7)} + \sqrt{48}\sqrt{(x-7)(x+7)} + \sqrt{56}\sqrt{(x-8)(x+7)} = 84 \end{equation*} is $91/11$. May someone offer an algebraic explanation to this?
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$(x-6)(x+7)+(x-8)(x+7)=2(x-7)(x+7)$ is maybe helpful. – Michael Hoppe Nov 01 '18 at 20:15
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Since $-1267/157$ and $91/11$ are the solutions, the corresponding quadratic is $1727x^2-350x-115297=0$. Hence there might be no elegant way to solve the equation. – Michael Hoppe Nov 01 '18 at 20:45
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One brute force approach is to subtract $84$ from both sides, then multiply both sides by a rationalizing factor to obtain a polynomial with integer coefficients (only the leading term's coefficient and the constant term need to be explicitly determined), and finally apply the rational root test. For a rationalizing factor (see here), use $(a+b+c+d)(a+b-c+d)(a+b-c-d)$ times $(a-b+c+d)(a-b+c-d)(a-b-c+d)(a-b-c-d),$ where $a^2=42(x-6)(x+7)$ and $b^2=48(x-7)(x+7)$ and $c^2=56(x-8)(x+7)$ and $d=84$ (I think). – Dave L. Renfro Nov 01 '18 at 20:45
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This 1850 paper by Arthur Cayley might be of help in simplifying some of the algebra (see the rationalized versions on p. 352 of the equation at the bottom of p. 351). Possibly @Michael Hoppe's observations could be of use in the clear-up process for obtaining the leading coefficient and constant term. – Dave L. Renfro Nov 01 '18 at 21:00
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Also, according to the bottom of the first page of this 1929 paper by Chaundy, rationalizing the equation $\sqrt{u+a}+\sqrt{u+b}+\sqrt{u+c}+\sqrt{u+d}=0$ results in an equation linear in $u,$ which along with @Michael Hoppe's second comment, suggests to me that rationalizing your equation might lead to a quadratic equation in $x.$ See also the solution to #169 on pp. 136-137 of Mathematical Gazette 1 #12 (October 1897). – Dave L. Renfro Nov 01 '18 at 21:13
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See also Article 114 (pp. 246-250) of this 1835 paper by Wilhelm August Förstemann and this 1852 paper by Arthur Cayley (another copy here), especially p. 98, on pp. 97-101 of Volume 8 of Cambridge and Dublin Mathematical Journal. Where did this equation come from? Surely you didn't just guess it's form. – Dave L. Renfro Nov 01 '18 at 21:29
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One may transform the equation to $$\sqrt{42}\sqrt{u}+\sqrt{24}\sqrt{u+w}+\sqrt{56}\sqrt{w} =84.$$ But how to rationalise this one? – Michael Hoppe Nov 02 '18 at 06:50
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@Michael Hoppe Why do you have $\sqrt{24}$? – user143462 Nov 02 '18 at 15:54
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@Dave L. Renfro The article by Chaundy and the solution to the problem posted in the Mathematical Gazette are interesting ... but I am not sure they are relevant to solving the radical expression posted by A Gal named Desire. – user143462 Nov 02 '18 at 16:14
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@user143462 It was $\sqrt{48}\sqrt{(u+w)/2}$ before. – Michael Hoppe Nov 02 '18 at 23:00
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@Dave L. Renfro You say to "multiply both sides by a rationalizing factor to obtain a polynomial with integer coefficients." What is the number? – A gal named Desire Nov 03 '18 at 14:51
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It is not a number but rather a (very lengthy) algebraic expression --- a polynomial with certain integer coefficients, to be more specific. I described the process in my answer to Rationalizing radicals (see also here, especially at the end), and I gave a worked out example in this 10 December 2005 sci.math post. Also useful is this example. – Dave L. Renfro Nov 04 '18 at 10:08
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@Dave L. Renfro These links discuss sums of integral multiples of radicals. The equation in my post has $\sqrt{42}$, $\sqrt{48}$, and $\sqrt{56}$. – A gal named Desire Nov 08 '18 at 16:21
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The terms can be rewritten so that each term is the square root of a non-radical expression. For example, $\sqrt{42}\sqrt{(x-6)(x+7)} = \sqrt{42(x-6)(x+7)}$ and $84 = \sqrt{7056}.$ – Dave L. Renfro Nov 08 '18 at 20:23