Question

If 2.5 is a root of the equation 2x^2+5x−q=0, what are the possible values of q?

Answer 1

since we know that x1=2.5

we can just just use vieta's theorem

x1+x2=-b/a

x1*x2=c/a

Answer 2

q = -25
`10 +5 = \pm √(5^2-4*2*q) \\ 225 = 25-8q \\\\8q = -200 \rightarrow q = -\frac {200}8 = -25`

Explanation:

Grab the generic solution of the equation.
`\frac 52 = (-5 \pm √(5^2-4*2*q))/(2*2)`. Let's isolate the radical, and square both sides.
`10 +5 = \pm √(5^2-4*2*q) \\ 225 = 25-8q \\\\8q = -200 \rightarrow q = -\frac {200}8 = -25`