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March 26, 2015

March 26, 2015

Posted by **GGIFT** on Monday, March 8, 2010 at 8:14am.

- ALGEBRA -
**Reiny**, Monday, March 8, 2010 at 8:39amWhen performing operations, like "squaring both sides", sometimes roots are introduced that were not part of the original relation.

e.g.

let x = 4, and only 4

square both sides

x^2 = 16

now take √

√(x^2) = √16

± x = 4

x = ±4 , we now a "new" root

- ALGEBRA -
**GGIFT**, Monday, March 8, 2010 at 8:56amSo inother words, we need to make that solutions are not those of original equations??

- ALGEBRA -
**Reiny**, Monday, March 8, 2010 at 9:26amI will give you another example

solve √(3x+10) = x+2

square both sides

3x+10 = x^2 + 4x + 4

x^2 + x - 6 = 0

(x+3)(x-2) = 0

x = -3 or x = 2

check: (in original)

if x=2

LS = √(6+10) = √16 = 4

RS = 2+2 = 4 , checks

if x = -3

LS = √(-9+1) = 1

RS = -3+2 = -1 , does not work

so x = 2

- ALGEBRA -
**GGIFT**, Monday, March 8, 2010 at 6:38pmI still do not understand it.

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