9514 1404 393
Answer:
x = √2
Step-by-step explanation:
A graph indicates the only solution is near x=√2.
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Square both sides, separate the radical and do it again.
[tex]\displaystyle(2-x)\sqrt{\frac{x+2}{x-1}}=\sqrt{x}+\sqrt{3x-4}\qquad\text{given}\\\\(2-x)^2\cdot\frac{x+2}{x-1}=x+(3x-4)+2\sqrt{x(3x-4)}\qquad\text{square}\\\\\frac{(2-x)^2(x+2)}{x-1}-4x+4=2\sqrt{x(3x-4)}\qquad\text{isolate radical}\\\\\left(\frac{(2-x)^2(x+2)-4(x-1)^2}{x-1}\right)^2=x(3x-4)\qquad\text{square}\\\\(x^3-6x^2+4x+4)^2=4(x-1)^2(3x^2-4x)\qquad\text{multiply by $(x-1)^2$}[/tex]
Now, we can put this polynomial equation into standard form and factor it.
[tex]x^6 -12x^5+32x^4-76x^2+48x+16=0\\\\(x-2)^2(x^2-2)(x^2-8x-2)=0\qquad\text{factor it}\\\\x\in \{2,\pm\sqrt{2},4\pm3\sqrt{2}\}[/tex]
The original equation requires that we restrict the domain of possible solutions. In order for the radicals to be non-negative, we must have x ≥ 4/3. In order for the left side of the equation to be non-negative, we must have x ≤ 2. So, the only potential solutions will be in the interval [4/3, 2].
The only values in the above list that match this requirement are {√2, 2}. We know that the right side of the equation cannot be zero, so the value x=2 is also an extraneous solution.
The only solution is x = √2.
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Additional comment
For solving higher-degree polynomials, I like to use a graphing calculator to help me find the roots. The second attachment shows the roots of the 6th-degree polynomial. They can help us factor the equation. (There are also various machine solvers available that will show factors and roots.)
