Question 1104240
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Since the problem specifies that x is not 0, we can simplify the form of the equation greatly by multiplying every term in the numerator and denominator of both fractions by {{{sqrt(x/2)}}}.  That will make the equation<br>
{{{(x+1-2sqrt(x))/(x+1) = (x+1)/(x+1+10sqrt(x))}}}<br>
Solving by "cross multiplying"...
{{{(x+1)^2 = (x+1)^2 + 8(x+1)*sqrt(x) - 20x}}}
{{{20x = 8(x+1)*sqrt(x)}}}
{{{5x = 2(x+1)*sqrt(x)}}}
{{{25x^2 = 4(x^2+2x+1)(x)}}}
{{{25x^2 = 4x^3+8x^2+4x}}}
{{{4x^3-17x^2+4x = 0}}}
{{{x(4x^2-17x+4) = 0}}}<br>
Since the problem asked for the product of the roots -- and not for the actual roots -- we can at this point say the product is 4/4 = 1, using the fact that the product of the roots of the quadratic equation ax^2+bx+c=0 is c/a.<br>
So the answer to the problem is: the product of the roots of the equation is 1.<br>
However, the quadratic expression factors nicely, making it easy to find the actual roots:<br>
{{{x(4x-1)(x-4) = 0}}}<br>
The roots are x=1/4 and x=4.