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3 years ago

10

Use square roots to solve theequation x2 = -121 over thecomplex numbers.

1 answer:

balandron [24]3 years ago

5 0

Answer:

x = ±11i

Step-by-step explanation:

x^2 = -121

Take the square root of each side

sqrt(x^2) = sqrt(-121)

sqrt(x^2) = sqrt (-1) *sqrt(121)

We know that the sqrt(-1) is i

x = i ( ±11)

x = ±11i

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Find the equation, (f(x) = a(x - h)2 + k), for a parabola containing point (2, -1) and having (4, -3) as a vertex. What is the s

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$f(x)=\frac{1}{2}x^2-4x+5$

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A parabola is written in the form

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where:

$h$ is the x-coordinate of the vertex of the parabola

$ak$ is the y-coordinate of the vertex of the parabola

$a$ is a scale factor

For the parabola in the problem, we know that the vertex has coordinates (4,-3), so we have:

$h=4$ (2)

$ak=-3$

From this last equation, we get that $a=\frac{-3}{k}$ (3)

Substituting (2) and (3) into (1) we get the new expression:

$f(x)=-\frac{3}{k}((x-4)^2+k) = -\frac{3}{k}(x-4)^2 -3$ (4)

We also know that the parabola contains the point (2,-1), so we can substitute

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Into eq.(4) and find the value of k:

$-1=-\frac{3}{k}(2-4)^2-3\\-1=-\frac{3}{k}\cdot 4 -3\\2=-\frac{12}{k}\\k=-\frac{12}{2}=-6$

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So the equation of the parabola is:

$f(x)=\frac{1}{2}((x-4)^2 -6)$ (5)

Now we want to rewrite it in the standard form, i.e. in the form

$f(x)=ax^2+bx+c$

To do that, we simply rewrite (5) expliciting the various terms, we find:

$f(x)=\frac{1}{2}((x^2-8x+16)-6)=\frac{1}{2}(x^2-8x+10)=\frac{1}{2}x^2-4x+5$

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3 years ago