As written, the "expression" (not "equation", since it contains no "equals" sign) is as follows:
. . . . .\(\displaystyle 16\, x^2\, +\, \dfrac{2}{3kx}\, +\, 9\)
However, I suspect that you meant to post "16x^2 + (2/3)kx + 9", which typesets as:
. . . . .\(\displaystyle 16\, x^2\, +\, \dfrac{2}{3}\, kx\, +\, 9\)
(Grouping symbols matter!)
I need to find the value of k that would make the equation a perfect square.
They've given you a formula:
. . . . .\(\displaystyle a^2\, x^2\, +\, 2\, a\, b\, x\, +\, b^2\, =\, (ax\, +\, b)^2\)
They've given you a quadratic with the first and third terms completed. Take the first term they gave you and the formula they gave you, and compute the value of "a". Take the third term they gave you and the formula they gave you, and compute the value of "b". Then take the values you've found for "a" and "b", along with the formula they gave you and the second term they gave you, and solve for the value of "k".
If you get stuck, please reply with a clear listing of your efforts in following the above steps. Thank you!
