Problem: If 4 x + 2 x = 1 4 x+\sqrt{2 x}=14 x + 2 x ​ = 1 , then x xx :
Answer Choices:
A. is an integer
B. is fractional
C. is irrational
D. is imaginary
E. may have two different values
Solution:
4 x + 2 x = 1 ∴ 4 x − 1 = − 2 x 4 x+\sqrt{2 x}=1 \quad \therefore 4 x-1=-\sqrt{2 x}4 x + 2 x ​ = 1 ∴ 4 x − 1 = − 2 x ​ \quad
16 x 2 − 10 x + 1 = 0 ∴ x = 1 8 16 \mathrm{x}^{2}-10 \mathrm{x}+1=0 \quad \therefore \mathrm{x}=\dfrac{1}{8}1 6 x 2 − 1 0 x + 1 = 0 ∴ x = 8 1 ​
x = 1 2 x=\dfrac{1}{2}x = 2 1 ​ does not satisfy the original equation.
or
Let y = x ∴ 4 y 2 + 2 y − 1 = 0 y=\sqrt{x} \quad \therefore 4 y^{2}+\sqrt{2} y-1=0y = x ​ ∴ 4 y 2 + 2 ​ y − 1 = 0 \quad
y = 2 2 8 , − 4 2 8 , x = y 2 = 1 8 , 1 2 y=\dfrac{2 \sqrt{2}}{8}, \quad \dfrac{-4 \sqrt{2}}{8}, x=y^{2}=\dfrac{1}{8}, \dfrac{1}{2}y = 8 2 2 ​ ​ , 8 − 4 2 ​ ​ , x = y 2 = 8 1 ​ , 2 1 ​ \quad