Find the value of \(x\):

\(x^2\) + (\(\frac {x} {x-1}\))\(^2\) = \(\frac {5} {4}\)

## Comments

Sort by:

TopNewest\( x = y + 1\)

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

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

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

\( a = y + \dfrac{1}{y}\)

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

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

\( a = - \dfrac{5}{2} , \dfrac{1}{2}\)

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

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

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

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

\( x = \dfrac{1}{2} , -3 , \dfrac{3 \pm i\sqrt{15}}{4}\) – U Z · 2 years, 7 months ago

Log in to reply

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. – Ninad Jadkar · 2 years, 7 months ago

Log in to reply

x= -1 or 1/2 – Ninad Jadkar · 2 years, 7 months ago

Log in to reply

– Mục Xiên · 2 years, 7 months ago

Can I know how to solve this, please?Log in to reply

Do you know how to clear denominators and factorize? – Calvin Lin Staff · 2 years, 7 months ago

Log in to reply

\(4x^2\) - \(8x^3\) + \(3x^2\) + \(10x\) - \(5\) = \(0\)

What to do next? – Mục Xiên · 2 years, 7 months ago

Log in to reply

rational root theorem

well, we want to start to factorize the polynomial. You can use theto guess at possible roots, and after which use the remainder factor theorem to help factorize. Have you seen these theorems before? – Calvin Lin Staff · 2 years, 7 months ago

Log in to reply

– Mục Xiên · 2 years, 7 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!Log in to reply