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What are the roots of this quadratic equation?

-10x^2 + 12x - 9 = 0

A. [tex]\( x = \frac{3}{5} \pm \frac{3i\sqrt{6}}{10} \)[/tex]

B. [tex]\( x = \frac{1}{5} \pm \frac{i\sqrt{6}}{5} \)[/tex]

C. [tex]\( x = -\frac{12}{5} \pm \frac{3i\sqrt{6}}{5} \)[/tex]

D. [tex]\( x = \frac{3}{10} \pm \frac{3i\sqrt{24}}{20} \)[/tex]


Sagot :

To solve the quadratic equation [tex]\(-10x^2 + 12x - 9 = 0\)[/tex], we will use the quadratic formula, which is:

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

where [tex]\(a = -10\)[/tex], [tex]\(b = 12\)[/tex], and [tex]\(c = -9\)[/tex].

First, we calculate the discriminant:

[tex]\[ \text{Discriminant} = b^2 - 4ac \][/tex]

Substituting the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:

[tex]\[ \text{Discriminant} = 12^2 - 4(-10)(-9) \][/tex]

[tex]\[ \text{Discriminant} = 144 - 360 \][/tex]

[tex]\[ \text{Discriminant} = -216 \][/tex]

Since the discriminant is negative ([tex]\(-216\)[/tex]), the roots will be complex numbers.

Next, we calculate the roots using the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{\text{Discriminant}}}{2a} \][/tex]

Substitute [tex]\(b\)[/tex], the discriminant, and [tex]\(a\)[/tex]:

[tex]\[ x = \frac{-12 \pm \sqrt{-216}}{2(-10)} \][/tex]

First, simplify [tex]\(\sqrt{-216}\)[/tex]:

[tex]\[ \sqrt{-216} = \sqrt{-1 \cdot 216} = i\sqrt{216} = i\sqrt{36 \cdot 6} = i \cdot 6 \cdot \sqrt{6} = 6i\sqrt{6} \][/tex]

Now we need to put it back into the formula:

[tex]\[ x = \frac{-12 \pm 6i\sqrt{6}}{-20} \][/tex]

Simplify the fractions by dividing the terms by the common factor:

[tex]\[ x = \frac{-12}{-20} \pm \frac{6i\sqrt{6}}{-20} \][/tex]

[tex]\[ x = \frac{3}{5} \pm \frac{3i\sqrt{6}}{10} \][/tex]

Thus, the roots of the equation [tex]\(-10x^2 + 12x - 9 = 0\)[/tex] are:

[tex]\[ x = \frac{3}{5} \pm \frac{3i\sqrt{6}}{10} \][/tex]

Therefore, the correct answer is:

A. [tex]\(\boxed{x = \frac{3}{5} \pm \frac{3 i \sqrt{6}}{10}}\)[/tex]