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If $p>0,\:q>0$, then the roots of $x^2-px-q=0$ are

(A) Imaginary.

(B) Both the roots are real and positive.

(C) Both the roots are real and negative.

(D) Both the roots are of opposite sign.

Can you answer this question?
 
 

1 Answer

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  • If the degree of a polynomial equation is even with constant term negative then the equation has atleast two real roots with opposite sign.
Given equation $x^2-px-q=0$ is of degree two.
$\therefore$ it has two real roots with opposite sign.
answered Jul 27, 2013 by rvidyagovindarajan_1
 

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