Show that the roots of the equation x^2+px+q=0 are rational,if p=k+q/k,where p,q,k are rational.

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For the roots of a  quadratic equation ax^2 + bx + c = 0 to be rational b^2 - 4ac >=0

We have the equation x^2 + px + q = 0

If p = k + q/k and the equation has rational roots:

p^2 - 4q

=> ( k...

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For the roots of a  quadratic equation ax^2 + bx + c = 0 to be rational b^2 - 4ac >=0

We have the equation x^2 + px + q = 0

If p = k + q/k and the equation has rational roots:

p^2 - 4q

=> ( k + q/k) - 4q

=> k^2 + q^2/k^2 + 2q - 4q

=> k^2 + q^2/k^2 - 2q

=> k^2 - 2q + q^2/k^2

=> (k - q/k)^2

As (k - q/k)^2 is a square it is greater than or equal to 0, which implies that  p^2 - 4q >=0.

Therefore the equation x^2+px+q = 0 has rational roots if p=k+q/k.

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