Question

Find the value for the quardic equation 2x^2-ax+64=0 so as one root is double the other​

Answer 1

Answer:

Let α and β are roots of the equation, 2x

2−ax+64=0

⇒Here, a=2,b=−a and c=64

According to the question,

⇒ β=2α

⇒ α.β= a.c

⇒ α.2α= 264

⇒ 2α 2 =32

⇒ α 2 =16

∴ α=±4

Put x=4

⇒ 2(4)

2 −a(4)+64=0

⇒ 4a=96

∴ a=24

Put x=−4

⇒ 2(−4)

2

−a(−4)+64=0

⇒ 4a=−96

∴ a=−24

∴ a=±24