Solving Quadratic Equations with Parameters | Step-by-Step Guide

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Homework Statement


x=t^2 +t
y=t^2-t


Homework Equations


quadratic equation


The Attempt at a Solution


0=t^2-t-x
0=t^2+t+y
<br /> 0=\frac{-1 +- \sqrt{1+4x}}{2}<br /> 0=\frac{1 +- \sqrt{1+4y}}{2}<br />

Aftert i set the to quadratics equal to one another, I do not know what to do. Any help?
 
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The results of the quadratic equations are t, not zero. But you are still doing it the hard way. Subtract the two original equations to get an expression for t in terms of x and y. Then substitute the resulting t back into one of the equations.
 
razored said:
x=t^2 +t
y=t^2-t

Hi razored! :smile:

Hint: go for the obvious … what is x + y? :wink:
 
Subtracting y from x leaves me with 0=x^2 -2xy+y^2-2y-2x which is ugly. Is there anything I can do to find to y= ? No, right?

Thank you.
 
What's to do about it? You eliminated the parameter. Isn't that what they asked for? Trying to solve for y will just make it super-ugly.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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