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

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Question

(B) Given the quadratic equation $2x^2 + 3x + 1 = 0$. Comparing this with $ax^2 + bx + c = 0$,we get $a = 2, b = 3, c = 1$. The discriminant $D = b^2

Vedclass · Accepted Answer

Rational

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

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

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