if one of the roots of the equation 2x^2+px+9=0is twice the other root,find the possible value of p.?

Question

please explain in details every solutions that you get.thank you.

please show me your working and explain it in details.thank you.

Answer 1

The fast method is to try and factor it by guessing. However, the most reliable method is to use the quadratic equation:
x = (-b +/- (b^2 - 4ac)^1/2) / 2a
x = (-p +/- (p^2 - 72)^1/2) / 4

Now simplify into an easier form:
x = -p/4 +/- (p^2 - 72)^1/2 / (16)^1/2
x = -p/4 +/- (p^2 - 72)^1/2 / (16)^1/2
x = -p/4 +/- ((p^2 - 72)/16)^1/2

Now define each root:
r1 = -p/4 + ((p^2 - 72)/16)^1/2
r2 = -p/4 - ((p^2 - 72)/16)^1/2

Since one root is twice the other, let r1 = 2r2:
r1 = 2r2
-p/4 + ((p^2 - 72)/16)^1/2 = 2(-p/4 - ((p^2 - 72)/16)^1/2)
-p/4 + ((p^2 - 72)/16)^1/2 = -2p/4 - 2((p^2 - 72)/16)^1/2
p/4 = -3((p^2 - 72)/16)^1/2
(p/4)^2 = (-3((p^2 - 72)/16)^1/2 )^2
p^2/16 = 9((p^2 - 72)/16)
p^2 = 9(p^2 - 72)
p^2 = 9p^2 - 648
-8p^2 = -648
p^2 = 81
p = 9

Now try the other way, where r2 = 2r1:
-p/4 - ((p^2 - 72)/16)^1/2 = 2(-p/4 + ((p^2 - 72)/16)^1/2)
-p/4 - ((p^2 - 72)/16)^1/2 = -2p/4 + 2(p^2 - 72)/16)^1/2
p/4 = 3(p^2 - 72)/16)^1/2
(p/4)^2 = (3((p^2 - 72)/16)^1/2 )^2
p^2/16 = 9((p^2 - 72)/16)

Note, that the step I stopped has an identical value to the previous attempt where r1 = 2r2, and we squared both sides. Thus, no matter how you determine the roots, you get the same answer for p.

Note, the first answerer was correct on the factoring, but got the wrong value of p. (2x + 3)(x+3) factors to 2x^2 + 9x + 9.

Answer 2

P could be 18 because 2x^2+18x+9 can be factored to (2x+3)(x+3), where -3 is twice as much as -1.5