The roots of the quadratic equation 2x^2 + 3x | vedclass

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(B) Given the quadratic equation: $2x^2 + 3x + 1 = 0$.Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 2$,$b = 3$,and $c = 1$.Th

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(B) Given the quadratic equation: $2x^2 + 3x + 1 = 0$.Comparing this with the standard form $ax^2 + bx + c = 0$,we have $a = 2$,$b = 3$,and $c = 1$.The discriminant $D$ is given by $D = b^2 - 4ac$.$D = (3)^2 - 4(2)(1) = 9 - 8 = 1$.Since $D > 0$ and $D$ is a perfect square,the roots are real and rational.Using the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$:$x = \frac{-3 \pm \sqrt{1}}{2(2)} = \frac{-3 \pm 1}{4}$.Thus,the roots are $x_1 = \frac{-3 + 1}{4} = \frac{-2}{4} = -\frac{1}{2}$ and $x_2 = \frac{-3 - 1}{4} = \frac{-4}{4} = -1$.Both roots,$-\frac{1}{2}$ and $-1$,are rational numbers.

The roots of the quadratic equation $2x^2 + 3x + 1 = 0$ are:

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