# Need some helps!

Find the value of $x$:
$x^2$ + ($\frac {x} {x-1}$)$^2$ = $\frac {5} {4}$ Note by Mục Xiên
5 years, 8 months ago

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I expanded it, and by hit and trial method, -1 was the solution. Then I divided the expanded polynomial by (x+1). The resultant cubic can be easily factorised.

- 5 years, 8 months ago

$x = y + 1$

$( y + 1)^2 + (\dfrac{y + 1}{y})^2 = \dfrac{5}{4}$

$y^2 + 2y + 1 + 1 + \dfrac{1}{y^2} + \dfrac{2}{y} = \dfrac{5}{4}$

$y^2 + \dfrac{1}{y^2} + 2y + \dfrac{2}{y} + 2 = \dfrac{5}{4}$

$a = y + \dfrac{1}{y}$

$a^2 - 2 + 2a + 2 = \dfrac{5}{4}$

$(a + 1)^2 = (\pm \dfrac{3}{2})^2$

$a = - \dfrac{5}{2} , \dfrac{1}{2}$

$y + \dfrac{1}{y} = - \dfrac{5}{2} , y + \dfrac{1}{y} = \dfrac{1}{2}$

$y = \dfrac{ -\dfrac{5}{2} \pm \sqrt{\dfrac{9}{4}}}{2} , y = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}$

$x - 1 = \dfrac{ -\dfrac{5}{2} \pm \sqrt{\dfrac{9}{4}}}{2} , x - 1 = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}$

$(x - 1= \dfrac{-1}{2} , -4) ~or~ x - 1 = \dfrac{ \dfrac{1}{2} \pm i\sqrt{\dfrac{15}{4}}}{2}$

$x = \dfrac{1}{2} , -3 , \dfrac{3 \pm i\sqrt{15}}{4}$

- 5 years, 8 months ago

x= -1 or 1/2

- 5 years, 8 months ago

Can I know how to solve this, please?

- 5 years, 8 months ago

What have you tried?

Do you know how to clear denominators and factorize?

Staff - 5 years, 8 months ago

I've already cleared denominators and got
$4x^2$ - $8x^3$ + $3x^2$ + $10x$ - $5$ = $0$
What to do next?

- 5 years, 8 months ago

well, we want to start to factorize the polynomial. You can use the rational root theorem
to guess at possible roots, and after which use the remainder factor theorem to help factorize. Have you seen these theorems before?

Staff - 5 years, 8 months ago

Well, I think I haven't seen these theorems before. Because I'm a middle school students. BTW, I'll ask my teacher for this problem's explaination, thanks for your helps!

- 5 years, 8 months ago