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timama [110]
3 years ago
7

What are the roots of the equation 16x2 + 40x + 27 = 0 in simplest a + bi form?

Mathematics
1 answer:
bezimeni [28]3 years ago
7 0

9514 1404 393

Answer:

  -5/4 +i(√2)/4 and -5/4 -i(√2)/4

Step-by-step explanation:

I find simplest form to be easier to get to if the leading coefficient is 1. Dividing by 16, we have ...

  x^2 +5/2x +27/16 = 0

Completing the square by adding and subtracting the square of half the x-coefficient, we get ...

  (x^2 +5/2x +25/16) +27/16 -25/16 = 0

  (x +5/4)^2 = 2/16

Subtracting 2/16, taking the square root, and subtracting 5/4 gives ...

  x +5/4 = ±√(-2/16)

  x = -5/4 ±i(√2)/4

The roots are -5/4 +i(√2)/4 and -5/4 -i(√2)/4.

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Part 1 -

1. f(x) = 5(x-2)^2 + 4

axis of symmetry: x = 2

vertex: (2, 4)

2. f(x) = 12(x + 6)^2 - 5

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vertex:  (-6, -5)

3. f(x) = 2x^2 + 8x - 7

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Y=x+3 and y=x-y help me ​
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Step-by-step explanation:

Given

y = x + 3.....equation 1

y = x - y

or y + y = x

or 2y = x

or y = x / 2......equation 2

Now

Putting equation 2 in equation 1

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x - x/2 = - 3

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