Question

solve 3√x-3-2=16 (radical ends above 3)

a.x=6
b.x=9
c.x=39
d.none of the above

solve √2x+15=√5(radical ends on top of 15)
a.x=10
b.x=5
c.x=-5
d.None of the above

rationalize the denominator of √-36 / (2-3i)+ (3+2i)

I would really appreciate the help!

Answer 1

For the first question, add 2 to both sides, what do you get?

Answer 2

\[3\sqrt{x+5}=16\] i think

Answer 3

Left side is correct, right side is not correct. You haven't add 2 to the right side

Answer 4

so would it be 18 instead of = 16?

Answer 5

=18 instead of = 16 :)
Here you got
\(3\sqrt{x+5}=18\)
Now, divide both sides by 3

Answer 6

well the 18 will be 6 but I don't know how to go about dividing the radical

Answer 7

Wait.... Not right...
It should be \(3\sqrt{x+3}=18\)

Answer 8

Sorry I overlooked something...
Once again.
Add 2 to both sides
\[3\sqrt{x-3}-2=16\]\[3\sqrt{x-3}-2+2=16+2\]\[3\sqrt{x-3}=18\]Divide both sides by 3.\[\frac{3\sqrt{x-3}}{3}=\frac{18}{3}\]Simplify it. What do you get from here?

Answer 9

would it be \[\sqrt{x}-1=6\]

Answer 10

Nope. You need not deal with the things in the square root for the moment. Just deal with the coefficient of the square root.

Answer 11

And you cannot break the square root in that way..

Answer 12

so \[\sqrt{x-3}=6?\]

Answer 13

Yes. Square both sides now. What do you get?

Answer 14

what do you mean by square them? I'm sorry I'm clearly really bad at math.

Answer 15

My bad :(
Take square on both sides.
\[\sqrt{x-3} ^2=6^2 \]Can you simplify it?

Answer 16

so I think either \[\sqrt{x^2-9}=36\] or\[\sqrt{x-9}=36\]

Answer 17

Nope. Things in the square root won't change.
\[\sqrt{x-3}^2=6^2\]\[x-3=36\]
Since = \(\sqrt a ^2 = a\)

Answer 18

so x=39 right?

Answer 19

Yes.

Answer 20

Thank you so much you helped me so much! thanks for being patient!

Answer 21

Welcome. question 2!
\[\sqrt{2x+15}=\sqrt{5}\]
Take square on both sides, what do you get?

Answer 22

So \[\sqrt{2x+15}^2=\sqrt{5}^2\]

Answer 23

Yes, and then?!

Answer 24

that removes the radicals right? so we'd get 2x+15=5

Answer 25

Yes. Next, subtract 15 from both sides.

Answer 26

so 2x=-10 which equals x=-5
Am I correct?

Answer 27

I think it's correct! :)

Answer 28

YYYYYYAYYYYYY! Thank you!

Answer 29

Welcome! Last one
\[\frac{-36}{ (2-3i)+ (3+2i)}\] ^Is this the question?

Answer 30

Yes that's the question

Answer 31

Uh-oh, or is it\[\frac{\sqrt{-36}}{ (2-3i)+ (3+2i)}\]

Answer 32

sorry yes its the second one

Answer 33

\(\sqrt{-1} = i\)
\(\sqrt{-36} = 6i\)

So, we can get \[\frac{\sqrt{-36}}{ (2-3i)+ (3+2i)} = \frac{6i}{ (2-3i)+ (3+2i)} \]Got it so far?

Answer 34

yes I get it so far

Answer 35

Hold on.. simplify the denominator first, can you?

Answer 36

how would I do that??

Answer 37

(2-3i) + (3+2i)
= (2+3) + (2i-3i)
= ...?

Answer 38

so 5+ -i

Answer 39

5+ (-i) = 5-i
So, you get \[\frac{\sqrt{-36}}{ (2-3i)+ (3+2i)}\]\[ = \frac{6i}{ (2-3i)+ (3+2i)}\]\[ = \frac{6i}{5-i}\]Now multiply the fraction by a conjugate \( \frac{5+i}{5+i}\)
\[ = \frac{6i}{5-i} \times \frac{5+i}{5+i}\]\[ = \frac{6i(5+i)}{(5-i)(5+i)} \]Can you simplify it?

Answer 40

Oh okay I get it so would it be \[-6+30i \over 26\]

Answer 41

You can further simplify it :)

Answer 42

how??

Answer 43

They have the common factor of 2.

Answer 44

\[2(-3+15i) \over 2\times13\]

Answer 45

\[-3+15i \over 13\] right ??

Answer 46

Yup.

Answer 47

Thank you big time! I truly appreciate your help and your time. Thank you again!