`x^(2)+2kx=(j)/(3)`
In the quadratic equation above, j and k are constant. What are the solutions of x ?

Options

A. `x=-kpm(sqrt(3(3k^(2)+j)))/(3)`
B. `x=-6kpm(sqrt(3(3k^(2)+j)))/(3)`
C. `x=-kpm(sqrt(3(3k^(2)+j)))/(6)`
D. `x=-6kpm(k+(sqrt(3j))/(3))`

Answer

A

Explanation

Difficult: Hard
Category: Passport to Advanced Math/ Quadratics
Strategy Advice: When an equation involves fractions, it is usually best to clear the fractions first to avoid messy calculations later. Also, when the answer choices involve square roots, you'll need to use the quadratic formula.
`x=(-bpmsqrt(b^(2)-4ac))/(2a)`
Getting to the Answer: Multiply both sides of the equation by 3 to eliminate the fraction on the right side, and then subtractt j from both sides to yield `3x^(2)+6kx-j=0`. This means `a=3, b=6k, and c=-j`. Substitute these values into the quadratic formula, and then simplify as shown below:
`x=(-(6x)pmsqrt((6x)^(2)-4(3)(-j)))/(2(3)) =(-6k)/(6)pm(sqrt(36k^(2)+12j))/(6) =-kpm(sqrt(4(9k^(2)+3j)))/(6) =-kpm(2sqrt(9k^(2)+3j))/(6) =-kpm(sqrt(3(3k^(2)+j)))/(3)`
Choice (A) is correct.