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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.

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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