The roots of the equation x^2 - sqrt{13}x + 1 | vedclass

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Question

(A) For a quadratic equation $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.Here, $a = 1$, $b = -\sqrt

Vedclass · Accepted Answer

Real and distinct

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(A) For a quadratic equation $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant $D = b^2 - 4ac$.Here, $a = 1$, $b = -\sqrt{13}$, and $c = 1$.Calculating the discriminant: $D = (-\sqrt{13})^2 - 4(1)(1) = 13 - 4 = 9$.Since $D > 0$ and $D$ is a perfect square, the roots are real, distinct, and rational if the coefficients are rational. However, since the coefficient $b = -\sqrt{13}$ is irrational, the roots are real and distinct.

The roots of the equation $x^2 - \sqrt{13}x + 1 = 0$ are:

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