Solve x^2 + 2kx + 3(k + 1) = 0 for K (using calculuator)

[chelser13]

I have been completely stumped by this problem:

x^2 + 2kx + 3(k+1) =0

Solve for K. K is a real positive number. Solve twice once with a radical and the second time with a calculator

 

[Loren]

Re: Algebra II problem with polynomials

x^2 + 2kx + 3(k+1) =0

Are you sure you are asked to sove for k? If you are solving for k...
Seems to me you should eliminate the parenthesis first, then gather all terms containing k on the left side and all other terms on the right side. Then factor out the k and do whatever is necessary so that you have k all by itself on the left side.

I think you will end up with k being equal to a fraction containing x's and whole numbers. I believe the fraction will be such that the numerator will always be less than zero (a negative number). That means if you want k to be a positive number, the denominator must be a negative number of some sort. I think you will end up with an answer such as k < something.

I don't see how that would involve a radical.

You will get a radical if you are supposed to solve for x. If I were solving for x I would involve the quadratic formula.

 

[Denis]

Re: Algebra II problem with polynomials

VERY UNCLEAR, but if you're looking for a solution where k is a positive integer: if x = -5, then k = 4

 

[stapel]

x^2 + 2kx + 3(k+1) =0

Solve for K. K is a real positive number.

If the instructions really told you to solve for "K", then you should point out to your instructor that there is no "K" in the equation, only a "k" (which is, of course, an entirely different variable). :shock:

Eliz.

 

[chelser13]

Re: Algebra II problem with polynomials

sorry i should have said that first you are to solve for x then solve for k

 

[stapel]

sorry i should have said that first you are to solve for x then solve for k

You have one equation with two variables. It is impossible to solve for a unique numerical solution. Were you perhaps supposed to solve for x "in terms of k", and then, starting over again at the beginning, solve for k "in terms of x"?

When you reply, please show all of your work and reasoning so far. Thank you!

Eliz.

 

[chelser13]

Re: Algebra II problem with polynomials

so far i have gotten x to one side and k on the other:

(x^2+3)+(2kx+3k)=0

 

[chelser13]

Re: Algebra II problem with polynomials

how do i solve x 'in terms of k' then solving k 'in terms of x'

 

[Denis]

Re: Algebra II problem with polynomials

so far i have gotten x to one side and k on the other:
(x^2+3)+(2kx+3k)=0

No: "each side" means each side of the equal sign.

Chelser, you seem to be quite lost...are you a student attending math classes?

 

[Deleted member 4993]

Re: Algebra II problem with polynomials

Use quadratic equation:

for

$\displaystyle A\cdot x^2 \, + \, B \cdot x \, + \, C = 0$

the solution for 'x' is:

$\displaystyle x_{1,2} = \frac {B^2 \, \pm \, \sqrt{B^2 \, - \, 4\cdot A \cdot C}}{2\cdot A}$