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Question

(A) Given the quadratic equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 3$,$b = 2\sqrt

Vedclass · Accepted Answer

$-\sqrt{5}, \frac{\sqrt{5}}{3}$

Vedclass · Answer

(A) Given the quadratic equation: $3x^{2} + 2\sqrt{5}x - 5 = 0$.Comparing this with the standard form $ax^{2} + bx + c = 0$,we get $a = 3$,$b = 2\sqrt{5}$,and $c = -5$.The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}$.First,calculate the discriminant $D = b^{2} - 4ac = (2\sqrt{5})^{2} - 4(3)(-5) = 20 + 60 = 80$.Now,substitute the values into the formula:$x = \frac{-2\sqrt{5} \pm \sqrt{80}}{2(3)} = \frac{-2\sqrt{5} \pm 4\sqrt{5}}{6}$.Case $1$: $x = \frac{-2\sqrt{5} + 4\sqrt{5}}{6} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}$.Case $2$: $x = \frac{-2\sqrt{5} - 4\sqrt{5}}{6} = \frac{-6\sqrt{5}}{6} = -\sqrt{5}$.Thus,the solutions are $-\sqrt{5}$ and $\frac{\sqrt{5}}{3}$.

Solve the following quadratic equation using the quadratic formula: $3x^{2} + 2\sqrt{5}x - 5 = 0$.

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