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elena55 [62]
3 years ago
6

What are the solutions of the quadratic equation below? 2x2 - 2x - 9 = 0

Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0

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

2x ^ 2-2x-9 = 0

The roots will come from:

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

Where:

a = 2\\b = -2\\c = -9

Substituting:

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}

The roots are:

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}

Answer:

Option C

Anettt [7]3 years ago
3 0

Answer: Option C

The solutions of the quadratic equation are:

x = \frac{1\±\sqrt{19}}{2}

Step-by-step explanation:

Use the quadratic formula to solve this equation.

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

x = \frac{-b\±\sqrt{b^2-4ac}}{2a}

In this case:

a=2\\b=-2\\c=-9

So

x = \frac{-(-2)\±\sqrt{(-2)^2-4(2)(-9)}}{2(2)}

x = \frac{1\±\sqrt{19}}{2}

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