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If $$x=y+\sqrt{y^2+1}$$ Then how to write $y$ in terms of $x$? Please tell me some hints

user602338
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4 Answers4

3

Hint: Just bring $y$ to the other side then square both sides.

$$x = y+\sqrt{y^2+1}$$

$$x-y = \sqrt{y^2+1}$$

$$(x-y)^2 = y^2+1$$

Here, note that $(a\pm b)^2 = a^2\pm 2ab+b^2$, and after necessary simplification, it should be clear enough.

KM101
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2

Hint. Note that $\text{arcsinh}(y)=\log(y+\sqrt{y^2+1})$. See Inverse hyperbolic sine.

Robert Z
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2

As $(\sqrt{y^2+1}-y)(\sqrt{y^2+1}+y)=\cdots=1,$

$\sqrt{y^2+1}+y=x\iff \sqrt{y^2+1}-y=\dfrac1x$

$x-\dfrac1x=?$

lab bhattacharjee
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  • See https://math.stackexchange.com/questions/400403/solve-sqrtx4-sqrtx1-1-for-x/400410#400410 https://math.stackexchange.com/questions/616536/how-to-solve-for-x-in-sqrt4x27-sqrt455-x-4/616537#616537 – lab bhattacharjee Dec 16 '18 at 17:11
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    (+1) This was the approach that came to mind when I saw the question. – robjohn Dec 16 '18 at 18:18
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I assume that

$x, y \in \Bbb R. \tag 0$

We are given

$x = y + \sqrt{y^2 + 1}. \tag 1$

We observe that this equation implies

$x > 0, \forall y \in \Bbb R, \tag{1.2}$

since

$y^2 < y^2 + 1 \Longrightarrow \vert y \vert < \sqrt{y^2 + 1}; \tag{1.5}$

we proceed:

$x - y = \sqrt{y^2 + 1}; \tag 2$

$x^2 - 2xy + y^2 = (x - y)^2 = y^2 + 1; \tag 3$

$x^2 - 2xy = 1; \tag 4$

$2xy = x^2 - 1; \tag 5$

by virtue of (1.2), we may divide by $2x$:

$y = \dfrac{x^2 - 1}{2x}. \tag 6$

CHECK:

From (6),

$y^2 = \dfrac{(x^2 - 1)^2}{4x^2} = \dfrac{1 - 2x^2 + x^4}{4x^2}; \tag 7$

$y^2 + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + 1 = \dfrac{1 - 2x^2 + x^4}{4x^2} + \dfrac{4x^2}{4x^2} = \dfrac{1 + 2x^2 + x^4}{4x^2} = \dfrac{(1 + x^2)^2}{4x^2}; \tag 8$

and again, by virtue of (1.2), we may apply $\sqrt \cdot$ to both sides and find

$\sqrt{y^2 + 1} = \dfrac{1 + x^2}{2x}; \tag 9$

$y + \sqrt{y^2 + 1} = \dfrac{x^2 - 1}{2x} + \dfrac{1 + x^2}{2x} = \dfrac{2x^2}{2x} = x. \tag{10}$

Robert Lewis
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