# Solving an equation with three quadratic radicals in the set of real numbers

How do we solve the following equation in the set of real numbers?

$$(26-x)\cdot\sqrt{5x-1} -(13x+14)\cdot\sqrt{5-2x} + 12\sqrt{(5x-1)\cdot(5-2x) }= 18x+32.$$

I tried putting $a=\sqrt{5x−1}$ and $b=\sqrt{5−2x}$ and then $2a^2+5b^2=23$.

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I tried put $a =\sqrt{5x-1}$ and $b = \sqrt{5-2x}$. And then $2a^2 + 5b^2=23.$ – minthao_2011 Sep 24 '12 at 2:11
$x=2$ looks like an answer. – NoChance Sep 24 '12 at 2:28
2 seems to work... – user133466 Sep 24 '12 at 2:41

If you put $a=\sqrt{5x-1}$ and $b = \sqrt{5-2x}$, the equation says $$(26-x) a - (13x+14) b + 12 a b = 18 x + 32$$ The resultant of $(26-x) a - (13x+14) b + 12 a b - 18 x - 32$ and $5x-1-a^2$ with respect to $a$ is $$-169\,{x}^{2}{b}^{2}+5\,{x}^{3}-588\,{x}^{2}b+356\,x{b}^{2}-585\,{x}^{ 2}+1808\,xb-340\,{b}^{2}+2280\,x-1520\,b-1700$$ The resultant of this and $5 - 2 x - b^2$ with respect to $b$ is $(49 x - 10)^3 (x - 2)^3$. So $x = 10/49$ or $2$. But now we have to check those by plugging in to the original equation. $x=2$ does work but $10/49$ doesn't work: in fact it would give you $$- \left( 26-x \right) \sqrt {5\,x-1}+ \left( 13\,x+14 \right) \sqrt {5 -2\,x}+12\,\sqrt { \left( 5\,x-1 \right) \left( 5-2\,x \right) }=18\, x+32$$

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Thank Robert Israel very much. – minthao_2011 Sep 24 '12 at 3:22

I solved by another way. This is my solution. Please comment to me. Put $t=\sqrt{5x-1}-\sqrt{5-2x}.$ We have $$t^2=3x+4-2\sqrt{(5x-1)( 5-2x)}$$ and $$t^3=(14-x)\sqrt{5x-1}-(13x+2)\sqrt{5-2x}.$$ And then, the given equation has the form $(t - 2)^3 = 0$, that is mean $t = 2$. With $t = 2,$ we have \begin{equation*} \sqrt{5x-1}-\sqrt{5-2x} = 2. \end{equation*} This equation has a solution $x =2.$ Thus, the given equation has the only root $x = 2.$

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Interesting method, I guess your substitution choice was lucky! – NoChance Sep 24 '12 at 9:51

I will continue where you left off.

$$a=\sqrt{5x−1}$$

We know that

$$\sqrt{5x−1} \geq 0$$

so

$$x \geq (1/5=0.2) ---- (Eq 1)$$

Given

$$2a^2+5b^2=23$$

Then

$$23-2a^2 \geq 0$$

Hence

$$\frac{23}{2} \geq a^2$$

That is:

$$\frac{23}{2} \geq 5x -1$$

Simplify to get:

$$x \leq 2.5$$

Combine this with (EQ 1) above, to get:

$$0.2 \leq x \leq 2.5$$

You may follow a similar process starting with $b$ to get a valid interval for $x$ and combing the two intervals you can get a close range of x. The problem can get simpler if you know that $x$ is an integer for example.

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This is another solution. Put $a=\sqrt{5x-1}$ and $b=\sqrt{5-2x}$. We have \begin{equation*} 26-x = \dfrac{47}{23}(5x - 1 ) + \dfrac{129}{23}(5-2x) = \dfrac{47}{23}a^2 + \dfrac{129}{23}b^2, \end{equation*} \begin{equation*} 13x+14= \dfrac{93}{23}(5x - 1 ) + \dfrac{83}{23}(5-2x) = \dfrac{93}{23}a^2 + \dfrac{83}{23}b^2, \end{equation*} \begin{equation*} 18x+32= \dfrac{154}{23}(5x - 1 ) + \dfrac{178}{23}(5-2x) = \dfrac{154}{23}a^2 + \dfrac{178}{23}b^2. \end{equation*} The given equation become \begin{equation*} \begin{cases} 47a^3+129ab^2-93ba^2-83b^3-154a^2-178b^2+276ab = 0,\\ 2a^2+5b^2=23. \end{cases} \end{equation*} This system of equations is solved here How do we solve the system of equations?

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