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

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(B) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.Here,$a = 1$,$b = -2\sqrt{2}$,and $c = 1$.$D = (-2

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Real and distinct

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(B) For the quadratic equation $ax^2 + bx + c = 0$,the discriminant $D$ is given by $D = b^2 - 4ac$.Here,$a = 1$,$b = -2\sqrt{2}$,and $c = 1$.$D = (-2\sqrt{2})^2 - 4(1)(1) = 8 - 4 = 4$.Since $D > 0$,the roots are real and distinct.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{2\sqrt{2} \pm \sqrt{4}}{2} = \frac{2\sqrt{2} \pm 2}{2} = \sqrt{2} \pm 1$.Since $\sqrt{2} \pm 1$ are irrational,the roots are real and distinct.

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

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