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aivan3 [116]
3 years ago
11

Solve the following quadratic equation using the quadratic formula.

Mathematics
2 answers:
IRINA_888 [86]3 years ago
8 0

x=4-3i/5 x=4+3-/5

plato users!!!!

sdas [7]3 years ago
6 0
ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a};\ x_2=\dfrac{-b+\sqrt\Delta}{2a}

<span>5x^2-8x+5=0\to a=5;\ b=-8;\ c=5\\\\\Delta=(-8)^2-4\cdot5\cdot5=64-100=-36\\\\\sqrt\Delta=\sqrt{-36}=6i

x_1=\dfrac{-(-8)-6i}{2\cdot5}=\dfrac{8-6i}{10}=\dfrac{4-3i}{5}\\\\x_2=\dfrac{-(-8)+6i}{2\cdot5}=\dfrac{8+6i}{10}=\dfrac{4+3i}{5}
</span>
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The function f(t) = t2 + 4t − 14 represents a parabola. Part A: Rewrite the function in vertex form by completing the square. Sh
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Answer:

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Given a parabola   f(t) = t² + 4 t − 14

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  Given a parabola   f(t) = t² + 4 t − 14

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                                 f¹(t) = 2 t + 4

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now                      t = \frac{-4}{2} = -2

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f(t) = (t +2)² - 18

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                   (x +2)² = y + 18

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The vertex ( h , k) = ( -2 , -18)

The Axis of the symmetry for f(t) is y -axis

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