Solve the following equation for u :
√(2u + 14) = √(3u -12)
where u is a real number.
(If there is more than one solution, separate them with commas.)
2007-05-10 19:38:47 · 5 answers · asked by AllStar Cheerleader 1 in Science & Mathematics ➔ Mathematics
square both sides of the eq to undo sq root
2u + 14 = 3u -12
-u + 14 = -12
-u = -26
u = 26
2007-05-10 19:42:53 · answer #1 · answered by Ana 4 · 0⤊ 0⤋
Square both sides: 2u + 14 = 3u - 12. It follows that u = 26. Plugging back into the original, we have sqrt(66) = sqrt(66), so we have one solution. By convention, the negative value of the square root is ignored.
2007-05-11 02:49:39 · answer #2 · answered by Anonymous · 0⤊ 0⤋
â(2u + 14) = â(3u -12)
We can take the whole equation to the second power, to leave the radicals out:
(â(2u + 14))^2 = (â(3u -12))^2
2u+14=3u-12
u=14+12
u=26
If we insert it back in the equation:
â(2u(26)+ 14) = â(3(26) -12)
â(52 + 14) = â(78 -12)
â(66) = â(66)
Which gives us a real, irrational, number...
2007-05-11 02:48:21 · answer #3 · answered by AV-2 1 · 0⤊ 0⤋
â(2u + 14) = â(3u -12)
Square both sides
2u+14=3u-12
Subtract 2u from both sides, add 12 to both sides.
u=26
Hope it's right and hope it helps!
2007-05-11 02:43:47 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Aquare both sides, giving you
2u + 14 = 3u - 12
u = 26
2007-05-11 02:43:06 · answer #5 · answered by Helmut 7 · 0⤊ 0⤋