Question

4x^2 - 16 = 0

2x^2 = 16

x^2 + 5 = 0
can someone plzz solve these equations for me plzzz

Answer 1

we are not here to do the work for you, but rather to guide you along the process

Answer 2

okay so help me

Answer 3

these are pretty basic, just one term containing the variable; algebra everything else to the other side of the equal sign; and sqrt

Answer 4

im still confused

Answer 5

how do you work an equation?

Answer 6

4u - 16 = 0, solve for u

Answer 7

subtract or add by both side idk

Answer 8

thats a good start yes :)

so lets add 16 to each side, then divide off that 4

Answer 9

okay let me try :)

Answer 10

so 4u=10 and then divide

Answer 11

4u - 16 = 0
+16 +16
------------
4u = 16

but yes, then divide by 4

Answer 12

lol sorry how did i get 10 i ment 16 sorry

Answer 13

so the answer is 4

Answer 14

the answer is almost ready :) one more thing to consider

Answer 15

so wat about 2x^2=16

Answer 16

yeah wat ?

Answer 17

instead of naming the unknown as "u", lets name it as "x^2"

x^2 = 4 ; now we to undo that ^2 part

Answer 18

take the square root of each side
\[\sqrt{x^2}=\pm\sqrt{4}\]
\[x=\pm2\]

Answer 19

okay i get that part

Answer 20

2x^2 = 16, dont let that x^2 scare you, just rename it if its bothering you: say x^2 = u

2u = 16 and solve for u

Answer 21

okay ley me try

Answer 22

i got 8

Answer 23

good, then:

u = 8 or rewritten in x^2 stuff: x^2 = 8, sqrt each side

Answer 24

is it \[\sqrt{x ^{2}}and \sqrt{2}\]

Answer 25

i see that your trying to look ahead perhaps :)
\[x^2 = 8\]
\[\sqrt{x^2} = \pm\sqrt{8}\]
\[x = \pm\sqrt{4*2}\]
\[x = \pm\sqrt{4}*\sqrt{2}\]
\[x = \pm2\sqrt{2}\]

Answer 26

the next one: x^2 + 5 = 0 works the same way, but the solution depends on what you are looking for.

Answer 27

lol oh okay let me try

Answer 28

so i can put it like 2u+5=o

Answer 29

x^2 + 5 = 0 ; let u = x^2 and rewrite

u + 5 = 0

Answer 30

so u is by it self or with the 5

Answer 31

theres a nice little trick i just saw, but it might confuse you a little: if we let x^2 = 5u we get:

5u + 5 = 0, solves to u=-1,

x^2 = 5(-1) = -5 is just as much a valid process as the rest of it

Answer 32

lol yeah it did kinda confused me

Answer 33

so do i solve it like that

Answer 34

lol, then lets just keep it as:\[x^2+5=0~:~let~x^2=u\]
\[u+5 = 0\\~~-5~~-5\\-----\\~u~~~~~ = -5\]
\[since~u=x^2, ~rewrite~as \]
\[x^2=-5\]\[\sqrt{x^2}=\pm\sqrt{-5}\]\[x=\pm\sqrt{-5}\]

Answer 35

have you come across anthing called imaginary numbers, or complex numbers?

Answer 36

um i dont think so

Answer 37

if you havent, then the answer to: x^2 + 5 = 0 is, no real solutions

Answer 38

the square root of a negative number has no real solutions; it only has complex (or imaginary) solutions

Answer 39

oh okay

Answer 40

i really need to practice more math is my weakness

Answer 41

practice does help :) good luck

Answer 42

thank u :)

Answer 43

can there be a square root for -25?