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How to solve this equation in the set real numbers $$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$ Using Mathematica, I know, this equation has two solutions $x = 1$ and $x = 3.$

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    $\begingroup$ Guessing solutions is not an option for you or? What is a solution for you? Only a analytical way, or are you interested in numerical solutions too ? $\endgroup$ Commented Mar 3, 2013 at 15:58

3 Answers 3

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I don't really recommend the following, but a general ansatz for $$\sqrt P - \sqrt Q = R$$ where $P$, $Q$, and $R$ are polynomials in a single unknown, might be to multiply with the conjugate: $$ P - Q = R\sqrt P + R\sqrt Q$$ then expand using $\sqrt P = R+\sqrt Q$ to get $$P - Q = R^2 + 2R\sqrt Q $$ and thus $$ (P - Q - R^2)^2 = 4R^2 Q$$ so the solutions you seek will be roots of some real polynomial -- but some of the roots will be non-solutions, and the final polynomial will generally have inconveniently high degree.

You can always do this and apply the rational root test to see if there are any nice solutions to the original equation, though. If not, then you might as well just solve the original numerically.

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Write:

$$\sqrt{8x + 1} - \sqrt{6x - 2} - 2x^2 + 8x - 7 = 0.$$

As:

$$(\sqrt{8x + 1} - \sqrt{6x - 2}) = (2x^2 - 8x + 7).$$

Squaring both sides yields:

$$14 x-2 \sqrt{6 x-2}\sqrt{8 x+1}-1 = 4x^4-32x^3+92x^2-112x+49$$

Simplifying:

$$-2 \sqrt{6 x-2}\sqrt{8 x+1} = 4x^4-32x^3+92x^2-112x+49$$

Squaring both sides again yields:

$$ 4(6x-2)(8x+1) = 16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+25076 x^2-12600 x+2500$$

Simplifying:

$$16 x^8-256 x^7+1760 x^6-6896 x^5+16928 x^4-26384 x^3+24884 x^2-12560 x+2508 = 0$$

This gives us more roots, so care must be taken, but indeed, we get back two of the $8$ roots as $x = 1$ and $x = 3$.

Numerical methods would also have worked earlier and guessing too.

Regards

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  • $\begingroup$ @ Amzoti If you don't use your computer, can you solve this way? $\endgroup$ Commented Mar 3, 2013 at 16:31
  • $\begingroup$ @minthao_2011: I would try guessing, looking for patterns... $\endgroup$
    – Amzoti
    Commented Mar 3, 2013 at 16:33
  • $\begingroup$ After finding $1$ and $3$ the remaining degree 6 polynomial turns out to be irreducible, so I won't try to compute $0.52485827544931491853347971043769008417$, $1.8557252925237647158115422848991734380$, $2.1338481507510021756733333358175364607$, $4.4517218947934982880461421562125906714$, $1.5169231932412099509677512563165046726\pm 1.8292676510086687936566891455556287623 i$ without computer. Not to mention check (the real numbers) if they fulfil the original equation - which they don't, as it turns out. $\endgroup$ Commented Mar 3, 2013 at 16:41
  • $\begingroup$ +1, this is slightly slicker than my suggestion -- specifically with respect to how easy it is to convince oneself the whole thing ought not to collapse to $0=0$ when simplifying. $\endgroup$ Commented Mar 3, 2013 at 17:28
  • $\begingroup$ @HenningMakholm: thank you and I always enjoy reading your answers as they help me learn! $\endgroup$
    – Amzoti
    Commented Mar 3, 2013 at 17:34
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Note that, with $x=1$, $y=\sqrt{8x+1}$ takes the value $y=3$ and with $x=3$, $y=\sqrt{8x+1}$ takes the value $y=5$. Put the point $A(1, 3)$ and $B(3,5)$. The equation of the line passing two points $A$ and $B$ is $y = x + 2.$ And then we write $$\sqrt{8x + 1}-(x+2).$$ Similar to with $$(x+1- \sqrt{6x - 2}).$$ We write the given equation has the form $$ \sqrt{8x + 1}-(x+2) + (x+1- \sqrt{6x - 2}) =2(x^2 - 4x + 3). $$ equavalent to $$ \dfrac{-(x^2 -4x + 3)}{ \sqrt{8x + 1}+(x+2)}+ \dfrac{x^2 -4x + 3}{x+1 + \sqrt{6x - 2}}=2(x^2 - 4x + 3).$$ Or $$ (x^2 -4x + 3)\left(\dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}}\right)=0.$$ With $x\geqslant \dfrac{1}{3}$, it is easy to see that $$ \dfrac{1}{ \sqrt{8x + 1}+(x+2)}+ 2- \dfrac{1}{x+1 + \sqrt{6x - 2}} >0.$$ Therefore we get $$ x^2 -4x + 3 = 0.$$

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  • $\begingroup$ What is so special about the equation of the secant line that it allowed you to factor so nicely? $\endgroup$
    – Ovi
    Commented Apr 24, 2013 at 5:57
  • $\begingroup$ You know for sure that the remaining function after you subtracted the secant term has roots at $1$ and $3$. $\endgroup$
    – user65203
    Commented Feb 7, 2014 at 7:12

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