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3 years ago

What indicators predict that a quadratic function will have a complex solution?

1 answer:

8 0

The nature of the roots can be determined by the determinant of the equation. The determinant is:
b² - 4ac

If this is positive, there are two roots
If this is 0, there is only one root
If this is negative, there are complex roots

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The graph of F(x), shown below, resembles the graph of G(x) = x2, but it has

mariarad [96]

Answer:F(x)=-x2-3

Step-by-step explanation:We are given a function:

The graph of is also shown in the given question figure.

It is a parabola with vertex at (0,0).

Sign of is positive, that is why the parabola opens up.

General equation of parabola is given as:

Here, in G(x), a = 1

Vertex (h,k) is (0,0).

As seen from the question figure,

The graph of F(x) opens down that is why it will have:

Sign of as negative. i.e.

And vertex is at (0,-3)

Putting the values of a and vertex coordinates,

Hence, the equation of parabola will become:

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2 years ago

How to find imaginary zeros anf real zeros of F(x)=-4x^5-8x^3+12x

Fofino [41]

Answer:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

Step-by-step explanation:

We are given the function:

$f(x)=-4x^5-8x^3+12x$

And we want to finds its zeros.

Therefore:

$0=-4x^5-8x^3+12x$

Firstly, we can divide everything by -4:

$0=x^5+2x^3-3x$

Factor out an x:

$0=x(x^4+2x^2-3)$

This is in quadratic form. For simplicity, we can let:

$u=x^2$

Then by substitution:

$0=x(u^2+2u-3)$

Factor:

$0=x(u+3)(u-1)$

Substitute back:

$0=x(x^2+3)(x^2-1)$

By the Zero Product Property:

$x=0\text{ and } x^2+3=0\text{ and } x^2-1=0$

Solving for each case:

$x=0\text{ and } x=\pm\sqrt{-3}\text{ and } x=\pm\sqrt{1}$

Therefore, our real and complex zeros are:

$x=\{0, -1, 1, -i\sqrt{3}, i\sqrt{3}\}$

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Answer:

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Step-by-step explanation:

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Answer:

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