Question

Pls help me,this is Canadian MO 1998 question,i'm very confuse with this​

Answer 1

The solution to the equation is x = 1.62

How to solve the equation?

The equation is given as:

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

Next, we split the equations as follows:

y = x

$y =\sqrt{x\ -\ \frac{1}{x}}+\sqrt{1-\ \frac{1}{x}}$

Next, we graph the equations (see attachment)

From the attached graph, the equations intersect at

(x, y)= (1.62, 1.62)

Remove the y value

x = 1.62

Hence, the solution to the equation is x = 1.62

Read more about equations at

brainly.com/question/2972832

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Answer 2

Rationalize the right side first. Expand and simplify.

$x = \sqrt{x - \dfrac1x} + \sqrt{1 - \dfrac1x}$

$x \left(\sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}\right) = \left(\sqrt{x - \dfrac1x} + \sqrt{1 - \dfrac1x}\right) \left(\sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}\right)$

$x \left(\sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}\right) = \left(\sqrt{x - \dfrac1x}\right)^2 - \left(\sqrt{1 - \dfrac1x}\right)^2$

$x \left(\sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}\right) = \left(x - \dfrac1x\right) - \left(1 - \dfrac1x\right)$

$x \left(\sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}\right) = x - 1$

Rearrange terms as

$\dfrac{x - 1}x = \sqrt{x - \dfrac1x} - \sqrt{1 - \dfrac1x}$

$\dfrac{x - 1}x + \sqrt{\dfrac{x - 1}x} = \sqrt{x - \dfrac1x}$

Substitute

$y = \sqrt{\dfrac{x-1}x} = \sqrt{1 - \dfrac1x} \\\\ \implies y^2 = 1 - \dfrac1x \implies x = \dfrac1{1-y^2}$

(noting that $y>0$) so that

$x - \dfrac1x = \dfrac1{1-y^2} - (1-y^2) = \dfrac{2y^2 - y^4}{1-y^2}$

and the equation transforms to

$y^2 + y = \sqrt{\dfrac{2y^2 - y^4}{1-y^2}}$

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

Solve for $y$.

$(y+1)^2 = \left(\sqrt{\dfrac{2-y^2}{1-y^2}}\right)^2$

$y^2 + 2y + 1 = \dfrac{2 - y^2}{1 - y^2}$

$-y^4 - 2y^3 + 2y + 1 = 2 - y^2$

$y^4 + 2y^3 - y^2 - 2y + 1 = 0$

$\left(y^4 + 2y^3 + y^2\right) - 2y^2 - 2y + 1 = 0$

$(y^2 + y)^2 - 2(y^2 + y) + 1 = 0$

$\left(y^2 + y - 1\right)^2 = 0$

$y^2 + y - 1 = 0$

$y = \dfrac{-1 \pm \sqrt5}2$

Since $y>0$, we take the positive root. Then

$y = \sqrt{\dfrac{x-1}x} = \dfrac{-1 + \sqrt5}2$

Solve for $x$.

$\left(\sqrt{\dfrac{x-1}x}\right)^2 = \left(\dfrac{-1 + \sqrt5}2\right)^2$

$\dfrac{x-1}x = \dfrac{6 - 2\sqrt5}4 = \dfrac{3 - \sqrt5}2$

$x - 1 = \dfrac{3-\sqrt5}2x$

$\dfrac{-1+\sqrt5}2 x = 1$

$x = \dfrac2{-1+\sqrt5}$

Rationalize to clean up the answer.

$x = \dfrac2{-1+\sqrt5}\cdot\dfrac{-1-\sqrt5}{-1-\sqrt5} = \boxed{\dfrac{1+\sqrt5}2} \approx 1.6108$

i.e. the golden ratio constant.