I tried to solving this it became 4th power of equationThe value of "x" in the equation x1−x+1−x=212

I am going to prove that the only real solution of the equation:x1−x+1−x=1is&nbsp;x=0.It is necessary that&nbsp;x∈[0,1]&nbsp;in order that the values inside the two square roots are nonnegative.Squaring both sides of the equation, we obtain:x1−x+1−x+2x=1x1−x−x+2x=0x−x+x21−x+2x=0x21−x+2x=0But&nbsp;x21−x≥0&nbsp;and&nbsp;&nbsp;2x≥0, so there is only a possibility which is:x21−x=2x=0Therefore&nbsp;x=0.As a matter of fact&nbsp;x=0&nbsp;is a real solution of the equation and it is actually the only one.

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High School
Algebra
Algebra I
Quadratic function and equation

Kennedy Ford

Answered question

2022-04-08

I tried to solving this it became 4th power of equation
The value of "x" in the equation

\frac{x}{1 - x} + \sqrt{1 - x} = 2 \frac{1}{2}

Answer & Explanation

madking711kzcb

Beginner2022-04-09Added 9 answers

I am going to prove that the only real solution of the equation:
$\sqrt{\frac{x}{1 - x}} + \sqrt{1 - x} = 1$
is $x = 0$ .
It is necessary that $x \in [0, 1]$ in order that the values inside the two square roots are nonnegative.
Squaring both sides of the equation, we obtain:
$\frac{x}{1 - x} + 1 - x + 2 \sqrt{x} = 1$
$\frac{x}{1 - x} - x + 2 \sqrt{x} = 0$
$\frac{x - x + x^{2}}{1 - x} + 2 \sqrt{x} = 0$
$\frac{x^{2}}{1 - x} + 2 \sqrt{x} = 0$
But $\frac{x^{2}}{1 - x} \geq 0 and 2 \sqrt{x} \geq 0$ , so there is only a possibility which is:
$\frac{x^{2}}{1 - x} = 2 \sqrt{x} = 0$
Therefore $x = 0$ .
As a matter of fact $x = 0$ is a real solution of the equation and it is actually the only one.

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