Question

find the solutions of the quadratic equation $4x^2 + 10x + 13 = 0$. choose 1 answer: a $-\frac{5}{4} pm \frac{3sqrt{3}}{4}i$ b $-\frac{5}{8} pm \frac{3sqrt{3}}{8}i$ c $\frac{5}{4} pm \frac{3sqrt{3}}{4}i$ d $-\frac{5}{4} pm \frac{3sqrt{3}}{4}$

Explanation:

Step1: Recall quadratic formula

For \(ax^2 + bx + c = 0\), solutions are \(x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 4\), \(b = 10\), \(c = 13\).

Step2: Calculate discriminant

Discriminant \(D=b^2 - 4ac=(10)^2 - 4\times4\times13 = 100 - 208=-108\).

Step3: Simplify square root of discriminant

\(\sqrt{D}=\sqrt{-108}=\sqrt{108}i = 6\sqrt{3}i\) (since \(\sqrt{-1}=i\)).

Step4: Apply quadratic formula

\(x=\frac{-10\pm6\sqrt{3}i}{2\times4}=\frac{-10\pm6\sqrt{3}i}{8}=\frac{-5\pm3\sqrt{3}i}{4}=-\frac{5}{4}\pm\frac{3\sqrt{3}}{4}i\).

Answer:

A. \(-\frac{5}{4}\pm\frac{3\sqrt{3}}{4}i\)