# Thread: finding the variables

1. ## finding the variables

what would be the varibles for:

-2x^2+6x-11=0
I know i use a quadratic formula but i get a negative square root and dont remember what to do next since its an imaginary number.

and

5 times 2^x+3=35
For this one i dont really get what to do first

2. Originally Posted by takkun0486
what would be the varibles for:

-2x^2+6x-11=0
I know i use a quadratic formula but i get a negative square root and dont remember what to do next since its an imaginary number.
Thus,
$\displaystyle x=\frac{-6\pm\sqrt{36-88}}{-4}$
Thus,
$\displaystyle x=\frac{-6\pm\sqrt{-52}}{-4}$
Thus,
$\displaystyle x=\frac{-6\pm 2i\sqrt{13}}{-4}$
Thus,
$\displaystyle x=\frac{3}{2}\pm \frac{\sqrt{13}}{2} i$
------
Ya got,
$\displaystyle 5(2^x+3)=35$
Divide by 5,
$\displaystyle 2^x+3=7$
Thus,
$\displaystyle 2^x=4$
Thus,
$\displaystyle 2^x=2^2$
Same bases thus,
$\displaystyle x=2$

3. Originally Posted by ThePerfectHacker
Thus,
$\displaystyle x=\frac{-6\pm\sqrt{36-88}}{-4}$
Thus,
$\displaystyle x=\frac{-6\pm\sqrt{-52}}{-4}$
Thus,
$\displaystyle x=\frac{-6\pm 2i\sqrt{13}}{-4}$
Thus,
$\displaystyle x=\frac{3}{2}\pm \frac{\sqrt{13}}{2} i$
------
Ya got,
$\displaystyle 5(2^x+3)=35$
Divide by 5,
$\displaystyle 2^x+3=7$
Thus,
$\displaystyle 2^x=4$
Thus,
$\displaystyle 2^x=2^2$
Same bases thus,
$\displaystyle x=2$
for the 2nd one i had the "x+3" as an exponent of 2

4. Originally Posted by takkun0486
for the 2nd one i had the "x+3" as an exponent of 2
Next time use parantheses,
Anyways the problem reduces to,
$\displaystyle 2^{x+3}=2^2$
Thus,
$\displaystyle x+3=2$
Thus,
$\displaystyle x=-1$

5. so it was
5(2^(x+3))=35

sry that i confused you on the other parts

6. any ideas?

7. Originally Posted by takkun0486
any ideas?
Oh!
Sorry I forgot you cannot subtract, then you have,
$\displaystyle 2^{x+3}=7$
Take logarithms,
$\displaystyle \log 2^{x+3}=\log 7$
Bring down exponent bases on rule,
$\displaystyle (x+3)\log 2=\log 7$
Thus,
$\displaystyle x+3=\frac{\log 7}{\log 2}$
Thus,
$\displaystyle x=\frac{\log 7}{\log 2}-3\approx -.19$