What is the product? 43 .4-3 0) 1) 4) 6)
2 answers:
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Answer: Either 
or 
depending on your teacher's preference. See the note at the bottom for more info.
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Let x be the speed limit. It is simply a placeholder. Think of it as a box with the number inside. We don't know what the number is, but we know that the largest it can be is 55. It cannot be larger than 55.
So x can be equal to 55 or it can be smaller
Put another way, x is less than or equal to 55 which is written as (on the keyboard you would type " x <= 55 " without quotes).
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Note: since negative speeds make no sense, we also must imply that x can't be negative. Your teacher may want you to write this in to clearly state it. If so, then you would also have
which when combined with the first inequality, you get this compound inequality 
(basically saying "x is some speed between 0 and 55 mph"). This very clearly states the boundaries on what x can be. Though your teacher may want you to stick to the first format as its simpler.
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Let x be the speed limit. It is simply a placeholder. Think of it as a box with the number inside. We don't know what the number is, but we know that the largest it can be is 55. It cannot be larger than 55.
So x can be equal to 55 or it can be smaller
Put another way, x is less than or equal to 55 which is written as
---------------------------------------
Note: since negative speeds make no sense, we also must imply that x can't be negative. Your teacher may want you to write this in to clearly state it. If so, then you would also have
Your answer is B, two complex roots and two real roots.
By factoring the original equation(which is a difference of two squares), you get:
Because first root is also a difference of two squares, it factors into x - 3 and x +3, your two real roots.
When you factor the second root, the roots are x - 3i and x + 3i.
To prove this, let's multiply them back together:
![[(x-3)(x+3)][(x-3i)(x+3i)]=0\\\\(x^{2}+3x-3x-9)(x^{2}+3xi-3xi-9i^{2})=0\\\\(x^{2}+0x-9)(x^{2}+0xi-9(-1))=0\\\\(x^{2}-9)(x^{2}+9)=0\\\\x^{4}+9x^{2}-9x^{2}-81=0\\\\x^{4}-81=0](https://tex.z-dn.net/?f=%5B%28x-3%29%28x%2B3%29%5D%5B%28x-3i%29%28x%2B3i%29%5D%3D0%5C%5C%5C%5C%28x%5E%7B2%7D%2B3x-3x-9%29%28x%5E%7B2%7D%2B3xi-3xi-9i%5E%7B2%7D%29%3D0%5C%5C%5C%5C%28x%5E%7B2%7D%2B0x-9%29%28x%5E%7B2%7D%2B0xi-9%28-1%29%29%3D0%5C%5C%5C%5C%28x%5E%7B2%7D-9%29%28x%5E%7B2%7D%2B9%29%3D0%5C%5C%5C%5Cx%5E%7B4%7D%2B9x%5E%7B2%7D-9x%5E%7B2%7D-81%3D0%5C%5C%5C%5Cx%5E%7B4%7D-81%3D0)
We reached the equation we started with, so that means that the roots are:
x + 3,
x - 3,
x + 3i, and
x - 3i,
two of which are real and two are complex.
By factoring the original equation(which is a difference of two squares), you get:
Because first root is also a difference of two squares, it factors into x - 3 and x +3, your two real roots.
When you factor the second root, the roots are x - 3i and x + 3i.
To prove this, let's multiply them back together:
We reached the equation we started with, so that means that the roots are:
x + 3,
x - 3,
x + 3i, and
x - 3i,
two of which are real and two are complex.