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- Thread starter nae99
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oh ok

- #28

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- #29

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x= -(-2) ± √ 4+12/ 2Now what's the answer?

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4

OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1

- #30

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What are you doing?x= -(-2) ± √ 4+12/ 2

x = -(2) [itex]\pm[/itex][itex]\sqrt{}[/itex] 16 [itex]/[/itex] 2

x = -(-2) [itex]\pm[/itex] 4 [itex]/[/itex] 2

x = -(-2) + 4 [itex]/[/itex] 2

x = 6[itex]/[/itex] 2

x= 4

OR

x = -(-2) - 2 [itex]/[/itex] 2

x = -2[itex]/[/itex] 2

x = -1

6/2=3 and 2-2=0.

Your both the answers are wrong. Correct them.

- #31

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OMG, such simple mistakes huh, okWhat are you doing?

6/2=3 and 2-2=0.

Your both the answers are wrong. Correct them.

x= 6/2 = 3

x= -(-2) - 2/ 2 = 0/2 x= 0

- #32

Mentallic

Homework Helper

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That's too much unnecessary work. From,

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).

- #33

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You're right Mentallic but nae99 is doing such silly mistakes here in this thread.

[tex](x-1)^2=4[/tex]

Take the square root of both sides (remember the [itex]\pm[/itex]).

@nae99- take care, since you need to consider this case:- x-1>0

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thanks much

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