Question 86697
SOLVE...
1. ROOT OF 4X+12-4=0
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{{{sqrt(4x+12)}}} - 4 = 0
{{{sqrt(4x+12)}}} = +4; add 4 to both sides
4x + 12 = 16; squared both sides, (gets rid of the radical)
4x = 16 - 12; subtract 12 from both sides
4x = 4
x = 1
Substitute 1 for x in the original equation and see that it checks out
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SIMPLIFY....
1. ROOT OF 50 OVER ROOT OF 2
:
{{{sqrt(50)/sqrt(2)}}}
{{{sqrt(25*2)/sqrt(2)}}}; find the perfect square inside the radical
{{{5*sqrt(2)/sqrt(2)}}}; extract the sqrt of 25
Ans = 5; The {{{sqrt(2)}}}'s cancel
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SIMPLIFY...
1. 6ROOT OF 6X+ 3 ROOT OF 6X
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{{{6*sqrt(6x) + 3*sqrt(6x)}}} = {{{9*sqrt(6x)}}}; sqrt(6x) are like terms, just add them
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EVALUATE IF POSSIBLE...
1.-3 ROOT OF 64
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{{{-3*sqrt(64)}}}: we know the sqrt(64) = 8
 -3 * 8 = -24
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did this help?