Compare using the quadratic formula to find solutions to a quadratic equation having irrational roots to that of one that has ra
1 answer:
For a general quadratic equation with rational coefficients
the two solutions are:
where D is the determinant.
Clearly, a solution will a rational number as long as is rational. However, it can be shown that a square root of an integer is only rational if its value is an integer. In other words,
is rational if and only if the determinant is a perfect square,
, otherwise the square root is irrational. Therefore the coefficients of quadratic equations that are to have rational solutions must satisfy the following condition:
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_______________
Evaluate the indefinite integral:
Make a trigonometric substitution:
so the integral (i) becomes
Now, substitute back for t = arcsin(x²), and you finally get the result:
✔
________
You could also make
x² = cos t
and you would get this expression for the integral:
✔
which is fine, because those two functions have the same derivative, as the difference between them is a constant:
![\mathsf{\dfrac{1}{2}\,arcsin(x^2)-\left(-\dfrac{1}{2}\,arccos(x^2)\right)}\\\\\\ =\mathsf{\dfrac{1}{2}\,arcsin(x^2)+\dfrac{1}{2}\,arccos(x^2)}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \left[\,arcsin(x^2)+arccos(x^2)\right]}\\\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{\pi}{2}}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29-%5Cleft%28-%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%5Cright%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carcsin%28x%5E2%29%2B%5Cdfrac%7B1%7D%7B2%7D%5C%2Carccos%28x%5E2%29%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cleft%5B%5C%2Carcsin%28x%5E2%29%2Barccos%28x%5E2%29%5Cright%5D%7D%5C%5C%5C%5C%5C%5C%0A%3D%5Cmathsf%7B%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%7D)
✔
and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.
I hope this helps. =)
_______________
Evaluate the indefinite integral:
Make a trigonometric substitution:
so the integral (i) becomes
Now, substitute back for t = arcsin(x²), and you finally get the result:
________
You could also make
x² = cos t
and you would get this expression for the integral:
which is fine, because those two functions have the same derivative, as the difference between them is a constant:
and that constant does not interfer in the differentiation process, because the derivative of a constant is zero.
I hope this helps. =)