x-x-x-x-m-x-x-x-m-is-a-real-parameter-

Question Number 98761 by bemath last updated on 16/Jun/20

\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x + \sqrt{x}}}

m is a real parameter

Commented by MJS last updated on 16/Jun/20

if x \in R

\Rightarrow x > 0 \land x - \sqrt{x} ⩾ 0 \Rightarrow x ⩾ 1

\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}} = m \sqrt{\frac{x}{x + \sqrt{x}}}

\sqrt{(1 + \sqrt{x}) \sqrt{x}} - \sqrt{(- 1 + \sqrt{x}) \sqrt{x}} = m \sqrt{\frac{\sqrt{x}}{1 + \sqrt{x}}}

x ⩾ 1 \Rightarrow we are allowed to divide by \sqrt{x}

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

1 + \sqrt{x} - \sqrt{- 1 + \sqrt{x}} \times \sqrt{1 + \sqrt{x}} = m

- 1 + \sqrt{x} ⩾ 0 \land 1 + \sqrt{x} ⩾ 0

1 + \sqrt{x} - \sqrt{x - 1} = m

\Rightarrow 1 < m ⩽ 2

\sqrt{x} - \sqrt{x - 1} = m - 1

squaring (both sides > 0)

2 x - 1 - 2 \sqrt{x (x - 1)} = {(m - 1)}^{2}

2 \sqrt{x (x - 1)} = 2 x - 1 - {(m - 1)}^{2}

to prove : 2 x - 1 - {(m - 1)}^{2} ⩾ 0

2 x - 1 ⩾ {(m - 1)}^{2}

x ⩾ 1 \Rightarrow 2 x - 1 ⩾ 1

\sqrt{2 x - 1} ⩾ m - 1

\sqrt{2 x - 1} + 1 ⩾ m

\sqrt{2 x - 1} + 1 ⩾ 2 \land m ⩽ 2 \Rightarrow true

squaring (both sides > 0)

4 x (x - 1) = {(2 x - 1 - {(m - 1)}^{2})}^{2}

4 {(m - 1)}^{2} x = {(m^{2} - 2 m + 2)}^{2}

x = \frac{{(m^{2} - 2 m + 2)}^{2}}{4 {(m - 1)}^{2}} \land 1 < m ⩽ 2

Commented by john santu last updated on 16/Jun/20

\sqrt{x + \sqrt{x}} {\sqrt{x + \sqrt{x}} - \sqrt{x - \sqrt{x}}} = m \sqrt{x}

x + \sqrt{x} - \sqrt{x^{2} - x} = m \sqrt{x}

x + (1 - m) \sqrt{x} = \sqrt{x^{2} - x}

squaring

x^{2} + 2 x \sqrt{x} (1 - m) + {(1 - m)}^{2} x = x^{2} - x

2 x \sqrt{x} (1 - m) + {(1 - m)}^{2} x + x = 0

x {2 \sqrt{x} (1 - m) + 2 - 2 m + m^{2}} = 0

2 \sqrt{x} = \frac{m^{2} - 2 m + 2}{m - 1} \Rightarrow \sqrt{x} = \frac{m^{2} - 2 m + 2}{2 m - 2}

x = {(\frac{m^{2} - 2 m + 2}{2 m - 2})}^{2}

Commented by MJS last updated on 16/Jun/20

your answer is only true for m = 2

Commented by john santu last updated on 16/Jun/20

typo sir . your answer and my answer

is same

Commented by MJS last updated on 16/Jun/20

I see . but we need to restrict m like I showed

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