Seudónimo Seudónimo
  • 08-04-2019
  • Mathematics
contestada

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

Respuesta :

Paounn
Paounn Paounn
  • 08-04-2019

Answer:

q = -25[tex]10 +5 = \pm \sqrt{5^2-4*2*q} \\ 225 = 25-8q \\\\8q = -200 \rightarrow q = -\frac {200}8 = -25[/tex]

Step-by-step explanation:

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

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ewash04 ewash04
  • 13-04-2019

Answer:

since we know that x1=2.5

we can just just use vieta's theorem

x1+x2=-b/a

x1*x2=c/a

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