Question

Find the real or imaginary solutions of each equation by factoring.

Answer 1

please help @allay

Answer 2

Allay, what's your issue?

Answer 3

my problem I don't understand

Answer 4

I'm going to need a little more info than that.

Answer 5

\[x ^{3}+27=0\]

\[-9x^{4}=-48x ^{2}+64\]

Answer 6

Apple wants your help to make a prototype for a larger iPod. They want to make the length of the new iPod Touch 1 ¼ times longer than it is now and the width 1 ½ times longer than it is now.
What will the dimensions (length and width) of the new iPod Touch prototype be, if the length and width are increased according to Apple’s specifications? Create a model showing the new dimensions.

4 25 inches x 2 13 inches
Above is NOT the dimensions for the prototype

Answer 7

Almost there, allay...

Answer 8

use this model

Answer 9

I'll be with you Taylor in just a moment.

Answer 10

no big deal, thanks

Answer 11

hello
@NickDantzlerward

Answer 12

Allay, here's how I did mess.

To find the length: (Four and one-fourth) + (Four and one-fourth) + seventeen-eighteenths

To find the width: (Two and thirteen-hundredths) + (Two and thirteen-hundredths) + (two-hundred-thirteen two-hundredths)

17/18 is 1/4 of (4 and 1/4).
213/200 is 1/2 of (2 and 13/100).

Add the given numbers to find the length, and add the other givens to find the width. You can simplify them into proper fraction form if you like.

Answer 13

Or turn them into decimals. Whatever.

Answer 14

Oooookay, Taylor...

Answer 15

I got length: 6 3/5
width: 3 1/2

Answer 16

I haven't simplified my work down, but that must be right then. Great job.

Answer 17

Taylor, to factor the first equation, isolate the 27. So like:

Instead of x^3 +27 = 0, subtract 27 from both sides and get: x^3 = -27.

We know that (-3)^3 = -27, so 'x' must be equal to -3.

Answer 18

I'll do the second one on paper and send you a pic.

Answer 19

so that would be a real solution? What's the difference between a real and imaginary solution?

Answer 20

It looks like surjithayer is about to give you a response. Hold on real quick, finishing the problem.

Answer 21

9x^4-48x^2+64=0
\[\left( 3x^2 \right)^2-2\left( 3x^2 \right)\left( 8 \right)+8^2=0\]
\[\left( 3x^2-8 \right)^2=0,3x^2=8,Repeated,x=\pm \sqrt{\frac{ 8 }{ 3 }}=\pm \frac{ 2\sqrt{2} }{ \sqrt{3}} Repeated.\]
these all are real solutions

Answer 22

\[x^3+27=0\]
\[x^3+3^3=0\]
\[a^3+b^3=\left( a+b \right)\left( a^2-ab+b^2 \right)\]

Answer 23

when there is negative sign under radical sign we get imaginary numbers.

Answer 24

Oh. He's got it covered...Alright then.

BTW, an imaginary number is i. We use it for cases in which we seem to have to find the root of say, -1. As we know, we can't find a root for any negative number. That's where 'i' comes in.

Answer 25

I'm confused. where did the 3x^2 come from??

Answer 26

9x^4=(3x^2)^2

Answer 27

\[\left( a-b \right)^2=a^2-2ab+b^2\]

Answer 28

what is that...

Answer 29

i wrote your second eq. by above formula.

Answer 30

\[x^3+27=\left( x+3 \right)\left( x^2-3x+9 \right)=0\]
either x+3=0
x=-3
or\[x^2-3x+9=0,x^2-3x+\left( \frac{ 3 }{ 2} \right)^2-\left( \frac{ 3 }{ 2 } \right)^2+9=0\]
\[\left( x-\frac{ 3 }{ 2 } \right)^2-\frac{ 9 }{ 4 }+9=0\]
\[\left( x-\frac{ 3 }{ 2 } \right)^2+\frac{ -9+36 }{ 4 }=0 \]
\[\left( x-\frac{ 3 }{ 2 } \right)^2+\frac{ 27 }{ 4 }=0 \]

Answer 31

\[(x-\frac{ 3 }{ 2 })^2-\left( \frac{ -{27} }{ 4 } \right)=0\]
\[\left( x-\frac{ 3 }{ 2 } \right)^2-(\frac{ \sqrt{27}\iota }{ 2 })^2=0\]
\[\left( x-\frac{ 3 }{ 2 }+\frac{ \sqrt{27}\iota }{ 2 } \right)\left( x-\frac{ 3 }{ 2 }-\frac{ \sqrt{27}\iota }{ 2 } \right) =0\]