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3 years ago

System of Linear Equations that have no solution? I need a system.

Mathematics

1 answer:

Bond [772]3 years ago

7 0

Answer:

A system of linear equations can have no solution, a unique solution or infinitely many solutions. ... for example 2x+3y=10, 2x+3y=12 has no solution. is the rref form of the matrix for this system.

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Which set of ordered pairs represents a function?

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2 years ago

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3 years ago

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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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2 years ago

4xsquared 12x 10 is less than or equal to 0 (or what multiplies to get 40 and adds to get 12)

g100num [7]

It either x=5 or x=-2 not sure which on

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2 years ago