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

Printable View

- May 23rd 2006, 06:57 PMtakkun0486finding 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 - May 23rd 2006, 07:04 PMThePerfectHackerQuote:

Originally Posted by**takkun0486**

$\displaystyle x=\frac{-6\pm\sqrt{36-88}}{-4}$

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,

$\displaystyle x=2$ - May 23rd 2006, 07:08 PMtakkun0486Quote:

Originally Posted by**ThePerfectHacker**

- May 23rd 2006, 07:13 PMThePerfectHackerQuote:

Originally Posted by**takkun0486**

Anyways the problem reduces to,

$\displaystyle 2^{x+3}=2^2$

Thus,

$\displaystyle x+3=2$

Thus,

$\displaystyle x=-1$ - May 23rd 2006, 07:16 PMtakkun0486
so it was

5(2^(x+3))=35

sry that i confused you on the other parts - May 23rd 2006, 07:24 PMtakkun0486
any ideas?

- May 23rd 2006, 07:28 PMThePerfectHackerQuote:

Originally Posted by**takkun0486**

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 $