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Find the value of \(x\):
\(x^2\) + (\(\frac {x} {x-1}\))\(^2\) = \(\frac {5} {4}\)

Note by Mục Xiên
2 years, 4 months ago

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\( x = y + 1\)

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

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

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

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

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

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

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

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

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

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

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

\( x = \dfrac{1}{2} , -3 , \dfrac{3 \pm i\sqrt{15}}{4}\) Megh Choksi · 2 years, 3 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. Ninad Jadkar · 2 years, 4 months ago

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x= -1 or 1/2 Ninad Jadkar · 2 years, 4 months ago

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@Ninad Jadkar Can I know how to solve this, please? Mục Xiên · 2 years, 4 months ago

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@Mục Xiên What have you tried?

Do you know how to clear denominators and factorize? Calvin Lin Staff · 2 years, 4 months ago

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@Calvin Lin I've already cleared denominators and got
\(4x^2\) - \(8x^3\) + \(3x^2\) + \(10x\) - \(5\) = \(0\)
What to do next? Mục Xiên · 2 years, 3 months ago

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@Mục Xiên 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? Calvin Lin Staff · 2 years, 3 months ago

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@Calvin Lin 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! Mục Xiên · 2 years, 3 months ago

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