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Algebra Difficulty 5.2 AIME, harder Find the answer

The article in the fifth issue of our journal in 1983, titled "A Simple Method for Compiling Radical Equations," states that the equation
5x1+2x=3x1 \sqrt{5 \mathrm{x}-1}+\sqrt{2 \mathrm{x}}=3 \mathrm{x}-1
"will produce a quartic equation after squaring twice, which may be quite troublesome to solve." In fact, this equation can be solved using a simpler method.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Solve the equation 3x1=(5x1)2x3 x-1=(5 x-1)-2 x
=(5x1+2x)=(\sqrt{5 x-1}+\sqrt{2 x})
- (5x12x)(\sqrt{5 x-1}-\sqrt{2 x})

and 5x1+2x>0\sqrt{5 x-1}+\sqrt{2 x}>0, from (1) we can get
5x12x=1 \sqrt{5 x-1}-\sqrt{2 x}=1 \text {. }

This indicates 5x11\sqrt{5 x-1} \geqslant 1, i.e., x25x \geqslant \frac{2}{5}.
 From (1)+(2) we get 25x2=3x,x259x220x+4=0,x25 \begin{array}{l} \text { From (1)+(2) we get } \\ 2 \sqrt{5 x-2}=3 x, \\ x \geqslant \frac{2}{5} \\ 9 x^{2}-20 x+4=0, \\ x \geqslant \frac{2}{5} \end{array}

We get x=2.(x=290)x=2 . \quad\left(x=\frac{2}{9}0\right),
Similarly, we can get (2), and solve x=2x=2.
From the above handling, we can summarize a unified solution method:
Let f(x),g(x),φ(x)f(x), g(x), \varphi(x) be polynomials:

The equation
f(x)±g(x)=v(x) \sqrt{f(x)} \pm \sqrt{g(x)}=v(x)
yields
f(x)g(x)=ψ(x). \sqrt{f(x)} \mp \sqrt{g(x)}=\psi(x) .

Solving (A) and (B) together, we can find f(x)\sqrt{f(x)} or g(x)\sqrt{g(x)}, and the equation can be simplified into a rational equation.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.