Respuesta :

carlosego

For this case we must find the solutions of the following quadratic equation:

[tex]2x ^ 2-2x-9 = 0[/tex]

The roots will come from:

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

Where:

[tex]a = 2\\b = -2\\c = -9[/tex]

Substituting:

[tex]x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (2) (- 9)}} {2 (2)}\\x = \frac {2 \pm \sqrt {4 + 72}} {2 (2)}\\x = \frac {2 \pm \sqrt {76}} {4}\\x = \frac {2 \pm \sqrt {2 ^ 2 * 19}} {4}\\x = \frac {2 \pm2 \sqrt {19}} {4}[/tex]

The roots are:

[tex]x_ {1} = \frac {2 + 2 \sqrt {19}} {4} = \frac {1+ \sqrt {19}} {2}\\x_ {2} = \frac {2-2 \sqrt {19}} {4} = \frac {1- \sqrt {19}} {2}[/tex]

Answer:

Option C

luisejr77

Answer: Option C

The solutions of the quadratic equation are:

[tex]x = \frac{1\±\sqrt{19}}{2}[/tex]

Step-by-step explanation:

Use the quadratic formula to solve this equation.

For a quadratic function of the form [tex]ax^2 +bx +c=0[/tex] the quadratic formula is:

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

In this case:

[tex]a=2\\b=-2\\c=-9[/tex]

So

[tex]x = \frac{-(-2)\±\sqrt{(-2)^2-4(2)(-9)}}{2(2)}[/tex]

[tex]x = \frac{1\±\sqrt{19}}{2}[/tex]

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