Please help me to solve these equations:
2 8
X = -
15
and:
2
X x 3x = -X + 4
Thank you !
please fix the format of your question. just use / to indicate fractions. example, type 1/x to mean $\displaystyle \frac 1x$ and 1/(x + 1) to mean $\displaystyle \frac 1{x + 1}$ etc, also, use ^ to indicate powers. type x^2 to mean $\displaystyle x^2$ for instance. you can use * to mean multiply. what you posted is confusing
Hello Shandy
I'm just guessing that your first equation might be
$\displaystyle x^2 = \frac{8}{15}$
If it is, the answers are
$\displaystyle x = \pm \sqrt{\frac{8}{15}}$
So get your calculator working! Divide 8 by 15. Then find the square root. Then put a $\displaystyle \pm$ sign in front, because there will be two answers.
The second one, I think, might be:
$\displaystyle x \times 3x = -x^2 + 4$
If so, then you solve it like this:
$\displaystyle x \times 3x = -x^2 + 4$
$\displaystyle \Rightarrow 3x^2 = -x^2 + 4$
$\displaystyle \Rightarrow 4x^2 = 4$
$\displaystyle \Rightarrow x^2 = 1$
$\displaystyle \Rightarrow x = \pm 1$
How am I doing? Did I guess right?
Grandad
Looking at your post inside "Quote" tags, some of the original formatting is displayed while in message-entry mode. Assuming you are using "X" and "x" to actually be the same variable, your equations appear to be:
. . . . .x^2 = 8/15
. . . . .(x)(3x) = -x^2 + 4
If so, then you can solve the first equation by taking the square root of each side of the equation, remembering the "plus-minus" on the right-hand side. You'll probably need to "rationalize the denominator" to get the answer in the "right" format.
For the second one, simplify the left-hand side to get 3x^2. Get the variable term together, and then divide 4x^2 = 4 on both sides by 4. Then you can solve by taking the square root of either side.
Have fun!