Using the square root property, solve for x
3x^2 + 2 = O
The book says the answer is plus/minus the (sq rt of 6)/3, but keep coming up with + or - sq. rt of (2/3). What am I doing wrong. Please show your work. Thanks!
2006-11-23 08:27:43 · 5 answers · asked by Inquisitor-2006 5 in Science & Mathematics ➔ Mathematics
Thanks, Stephen. My math tutor is away for the day! I'll be sure and send 10 points your way ASAP.
2006-11-23 08:42:01 · update #1
You're not doing anything wrong! Your answer and their answer are actually the same.
What do you get when you square sqrt(6)/3? You get 6/(3*3) = 6/9 = 2/3. So sqrt(6)/3 = sqrt(2/3).
I would prefer the answer you came up with, but theres really no difference.
2006-11-23 08:30:31 · answer #1 · answered by stephen m 4 · 0⤊ 0⤋
Hello,
The answers are the same. The book just rationalizes the denominator. It multiplies sqrt(2/3) by sqrt(3/3), which gives you sqrt(6)/3.
2006-11-23 08:38:44 · answer #2 · answered by toyallhi 2 · 0⤊ 0⤋
It is called rationalizing the denominator where you times the denominator by itself in order to remove any sqrts from the bottom
sqrt2/sqrt3
=sqrt2/sqrt3 * sqrt3/sqrt3
[as sqrt3/sqrt3 = 1 all you doing is multiplying by 1 so you are not changing the value]
=sqrt2*sqrt3/3
=(sqrt6)/3
The only problem with 3x^2 + 2 = O is that 3x^2=-2
x^2=-2/3
but sqrt of a negative number is not possible unless you know about i and complex numbers.
I presum you meant to write 3x^2 - 2 = O
2006-11-23 08:37:58 · answer #3 · answered by Oz 4 · 1⤊ 0⤋
3x^2+2=0
3x^2=-2
x^2=-2/3
x=+/-sqrt(2/3) i=+/-sqrt(6)/3 i
sqrt (1/3)=sqrt(3/9)=sqrt(3) /3
it is traditional to have all the radicals in the numerator.
Both you & the book missed the "i" (sqrt -1)
2006-11-23 08:43:59 · answer #4 · answered by yupchagee 7 · 0⤊ 0⤋
If you are to solve this over the set of real numbers, then the solution is the empty set.
2006-11-23 18:54:52 · answer #5 · answered by Jerry M 3 · 0⤊ 0⤋