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In simplest radical form, what are the solutions to the quadratic equation [tex]0 = -3x^2 - 4x + 5[/tex]?

Quadratic formula: [tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]

A. [tex]x = -\frac{2 \pm \sqrt{19}}{3}[/tex]
B. [tex]x = -\frac{2 \pm 2\sqrt{19}}{3}[/tex]
C. [tex]x = \frac{2 \pm \sqrt{19}}{3}[/tex]
D. [tex]x = \frac{2 \pm 2\sqrt{19}}{3}[/tex]


Sagot :

To find the solutions to the quadratic equation [tex]\(0 = -3x^2 - 4x + 5\)[/tex] using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], follow these steps:

1. Identify the coefficients:\
The quadratic equation is given in the standard form [tex]\(ax^2 + bx + c = 0\)[/tex]. Here:
[tex]\[ a = -3, \quad b = -4, \quad c = 5 \][/tex]

2. Calculate the discriminant ([tex]\(\Delta\)[/tex]):
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = (-4)^2 - 4(-3)(5) = 16 + 60 = 76 \][/tex]

3. Express the solutions using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substitute [tex]\(b\)[/tex], [tex]\(\Delta\)[/tex], and [tex]\(a\)[/tex]:
[tex]\[ x = \frac{-(-4) \pm \sqrt{76}}{2(-3)} \][/tex]
Simplify the numerator:
[tex]\[ x = \frac{4 \pm \sqrt{76}}{-6} \][/tex]

4. Simplify the expression under the square root ([tex]\(\sqrt{76}\)[/tex]) as needed:
[tex]\[ \sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19} \][/tex]
So, we have:
[tex]\[ x = \frac{4 \pm 2\sqrt{19}}{-6} \][/tex]
Simplify by dividing both the numerator terms by the denominator:
[tex]\[ x = -\frac{4 \pm 2\sqrt{19}}{6} \][/tex]
Further simplify by factoring out the common terms in the numerator and denominator:
[tex]\[ x = -\frac{2(2 \pm \sqrt{19})}{6} = -\frac{2}{6} (2 \pm \sqrt{19}) = -\frac{1}{3} (2 \pm \sqrt{19}) \][/tex]

Therefore, the solutions in simplest radical form are:
[tex]\[ x = -\frac{2 \pm \sqrt{19}}{3} \][/tex]

Thus, the correct answer is:
[tex]\[ x = -\frac{2 \pm \sqrt{19}}{3} \][/tex]
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