Question

One root of the equation, 5x^2 + 4x = k (for some real k) is 2, what is the other?

Answer 1

for the polynomial: \[ax^2 + bx + c = 0\]\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
there are 2 answer for x. one for the plus and one for the minus in the plus/minus sign. for the polynomial you gave:
a = 5
b = 4
c = -k

since x is 2 and b is 4, we know that we have to use the PLUS on the plus/minus sign. 2 > -4

plugging this in:\[2 = \frac{ -4 + \sqrt{4^2 - 4(5)(-k)} }{ 2(5) } = \frac{ -4 + \sqrt{16 + 20k} }{ 10 }\]\[20 = -4 + \sqrt{16 + 20k}\]\[24 = \sqrt{16 + 20k} \]\[24^2 = 16 + 20k\]\[k = 28\]

now that you have defined values for a, b and c. plug them in the quadratic equation to solve for x using the MINUS sign on the plus/minus sign