Question

Use the quadratic formula to solve for x:
x+1/x = x/2 I'll give you a medal ;)

Answer 1

\[- (1/2) x = 1/x\]
\[-(1/2) x^{2} -1 = 0\]
\[x = \frac{ -b +/- \sqrt{b^{2} - 4ac} }{ 2a }\]
\[x= \frac{ 0 +/- \sqrt{0 - 2}}{ -1 }\]

Answer 2

the answer in the book is \[ 1+/-\sqrt{3}\]

Answer 3

but I don't know how they got that -.- and the answer you've got is totally different...

Answer 4

is the equation \[x + \frac{ 1 }{ x } = \frac{ x }{ 2 }\]

Answer 5

No it's \[\frac{ x+1 }{ x } = \frac{ x }{ 2 }\]

Answer 6

\[x*x/(x+1) = 2\]
\[x^{2}=2x+2\]
\[x^{2}-2x-2 = 0\]

Answer 7

to do this you first put the equation in standard form:
\[ \frac{x+1}{x}= \frac{x}{2}\]
you put it in standard form by getting rid of the fractions. multiply both sides by x
\[ x\cdot \frac{x+1}{x}= x \cdot \frac{x}{2}\]
which simplifies
\[ \cancel{x}\cdot \frac{x+1}{\cancel{x}}=\frac{x^2}{2}\]
now multiply both sides by 2
\[ 2(x+1)= 2 \cdot \frac{x^2}{2}\]
cancel the 2's on the right side
\[ 2(x+1)= \cancel{2} \cdot \frac{x^2}{\cancel{2}}\]
on the left side, distribute the 2 by multipying by each term inside the parens
\[ 2x+2 = x^2\]
add -2x -2 to both sides
\[ 2x+2 -2x-2= x^2-2x-2\]
finally we get
\[ x^2-2x-2 =0\]

now apply the quadratic formula
see http://www.khanacademy.org/math/algebra/quadratics/v/applying-the-quadratic-formula