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solve for x where x is a real number.

2007-09-23 14:09:28 · 5 answers · asked by MTak 2 in Science & Mathematics Mathematics

5 answers

squaring both sides,we get
3x-9=2x+2
3x-2x=2+9
x=11

2007-09-23 14:16:28 · answer #1 · answered by alpha 7 · 0 0

if you square both sides of the equation you get:

3x - 9 = 2x +2
3x-2x = 2+9
x = 11

therefore sq rt of 33-9= sq rt of 22+2
therefore sq rt of 24= sq rt of 24

since the numbers under the square root are not negative the answer works

2007-09-23 21:16:13 · answer #2 · answered by CJ 3 · 0 0

sq rt of 3x-9= sq rt of 2x+2
square both sides

3x - 9 = 2x + 2
x = 11

Check
√(33 - 9) = √(22 + 2)

√24 = √24

2007-09-23 21:15:43 · answer #3 · answered by Marvin 4 · 1 0

Since both are under the square root, that simply means both expressions under the square root must be greater than or equal to zero. This only becomes important when checking your result. Other than that, they can be ignored by squaring both sides.

3x - 9 = 2x + 2

x = 11

Since 3*11-9 > 0, and 22 + 2 > 0, the answer x = 11 is acceptable.

2007-09-23 21:15:54 · answer #4 · answered by lhvinny 7 · 1 0

Sqrt(3x-9)=Sqrt(2x+2)

Square both sides:

(Sqrt(3x-9))^2=(Sqrt(2x+2))^2

3x-9=2x+2

Add 9 to both sides:

3x=2x+11

Subract 2x from both sides

x=11

2007-09-23 21:20:06 · answer #5 · answered by bri 1 · 0 0

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