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by Inactive member. on November 8th, 2009
I lost you. How many problems are you posting here?
Math question found in +venture in math (s'pore)
by HANz318 on October 30th, 2011
| 1 person likes this
Would you watch a video called 2 elephants 1 cup?
by randylikesyourbutt on October 28th, 2011
| 1 person likes this
use the factor theorem to determine whether or not h(x) is a factor of f(x)
h(x) = x + 1; f(x)= x^3 - 4x^2 + 3x + 8
by Anonymous on September 27th, 2011
| 2 people like this
what is the vertex of thsi problem!! g(x) =x^2+8x-1
by Anonymous on September 27th, 2011
| 1 person likes this
Math homework help please?
by Anonymous on September 28th, 2011
| 1 person likes this
You're reading How do you solve the following questions??? I've been trying for days. Please help me!! 3. (x^2 + 3)^2 - 6(x^2 + 3) = 7 7. 3*(the sq. root of) -384 30. x^2 + 14 = 0 6. (2/x)-(5/x)=1
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Comments
I think you have 3 problems. I'm not sure about the first two, but here's the third one.
(2/x)-(5/x)=1
This is the same as (2-5)/x =1
because they have the same denominators. Therefore...
-3/x=1
then multiply x on both sides, now you get...
-3 = x*1
-3 = x ...same as... x = -3
by Inactive member. on November 8th, 2009
Here are my attempts to the first two...
(x sqr + 3) sqr - 6(x sqr +3) = 7
if you expand the left side of the eqn, you get:
x to the 4th power + 3x sqr + 3x sqr + 9 - 6x sqr - 18 = 7
{3x sqr + 3x sqr cancels with 6x sqr), thus you now have:
x to the 4th power + 4 - 18 = 7, now solve for x:
x to the 4th power = 7 - 4 + 18
x to the 4th power = 21
There is no such thing as a square root of a negative number. It would have to be classified as an imaginary number, designated by the letter: i. If that's the case, then the eqn would be 3 * sqr root of 384 * sqr root of i. Which would be 58.79 times sqr root of i.
If the other equation is x sqr + 14 = 0, then there is no solution. Becasue x sqr cannot be equal to -14 (or any negative number for that matter). Unless again, if you're also dealing with imaginary numbers.
Hope this helped and didn't confuse you more.
by Inactive member. on November 8th, 2009