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Katyanochek1 [597]
3 years ago
7

Write x2 - 2x + 5 = 0 in vertex form. (Be sure to write your response in the form a(x-h)^2+k=0 with no spaces!)

Mathematics
1 answer:
Drupady [299]3 years ago
4 0
To write the given quadratic equation to its vertex form, we first form a perfect square.
 
                x² - 2x + 5 = 0
Transpose the constant to other side of the equation,
                x² - 2x = -5
Complete the square in the left side of the equation,
              x² - 2x + (-2/1(2))² = -5 + (-2/1(2))² 
Performed the operation,
                x² - 2x + 1 = -5 + 1
Factor the left side of the equation,
               (x - 1)² = -4

Thus, the vertex form of the equation is,
          <em>  (x-1)² + 4 = 0</em>
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Solve using the quadratic formula 3x^2+11x+5=0
storchak [24]

Answer:

3x^2 + 11x + 5 = 0 : x = \frac{-11 + \sqrt{61} }{6} , x = \frac{-11 - \sqrt{61} }{6}

Decimal:

(x = -0.53162..., x = -3.13504...)

Step-by-step explanation:

3x^2+11x+5=0

Solve with the quadratic formula:

For a quadratic equation of the form ax^2+bx+c=0 the solutions are:

x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

For a=3,\:b=11,\:c=5:\quad x_{1,\:2}=\frac{-11\pm \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3}

x=\frac{-11 + \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 + \sqrt{61} }{6}

x=\frac{-11 - \sqrt{11^2-4\cdot \:3\cdot \:5}}{2\cdot \:3} : \frac{-11 - \sqrt{61} }{6}

The solutions to the quadratic equation are:

x=\frac{-11+\sqrt{61}}{6},\:x=\frac{-11-\sqrt{61}}{6}

Hope I helped. If so, may I get brainliest and a thanks?

Thank you, have a good day! =)

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