Question

\[ x^{2}+2 k x=\frac{j}{3} \] In the quadratic equation above, \( j \) and \( k \) are constants. What are the solutions for \( x \) ? A) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \) B) \( x=-6 k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \) C) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{6} \) D) \( x=-6 k \pm\left(k+\frac{\sqrt{3 j}}{3}\right) \)

          \[
x^{2}+2 k x=\frac{j}{3}
\]

In the quadratic equation above, \( j \) and \( k \) are constants. What are the solutions for \( x \) ?
A) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \)
B) \( x=-6 k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \)
C) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{6} \)
D) \( x=-6 k \pm\left(k+\frac{\sqrt{3 j}}{3}\right) \)

x^2+2 k x=(j)/(3)

In the quadratic equation above, j and k are constants. What are the solutions for x ?
A) x=-k ±(√(3(3 k^2+j)))/(3)
B) x=-6 k ±(√(3(3 k^2+j)))/(3)
C) x=-k ±(√(3(3 k^2+j)))/(6)
D) x=-6 k ±(k+(√(3 j))/(3))

Added by Tammy J.

Engineering Mathematics

K. A. Stroud, Dexter J. Booth 5th Edition

Chapter 16

Instant Answer

Solved by Expert Rylie Howey

Step 1

Step 1: Start with the given equation: \[ x^2 + 2kx = \frac{j}{3} \] Show more…

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\[ x^{2}+2 k x=\frac{j}{3} \] In the quadratic equation above, \( j \) and \( k \) are constants. What are the solutions for \( x \) ? A) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \) B) \( x=-6 k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{3} \) C) \( x=-k \pm \frac{\sqrt{3\left(3 k^{2}+j\right)}}{6} \) D) \( x=-6 k \pm\left(k+\frac{\sqrt{3 j}}{3}\right) \)