Question

If x is real and p = (3(x^2 + 1) / (2x - 1) , prove that p^2 - 3(p + 3) >= 0

Answer 1

if: p = (3(x^2 + 1) / (2x - 1)
therefore
p^2 - 3(p + 3) = (9 ( -x^2 + x + 1)^2)/(1 - 2x)^2

Answer 2

= (9x^4 - 18x^3 - 9x^2 + 18x + 9)
---------------------------------
(4x^2 - 4x + 1)

Answer 3

\[p^2 - 3(p +3) >= 0\]

Answer 4

\[p = (3(x^2 + 1) / (2x -1))\]

Answer 5

sure, so when does
\[\frac{ (9x^4 - 18x^3 - 9x^2 + 18x + 9) }{ (4x^2 - 4x + 1) } = 0or +ve\]

Answer 6

how did you do that?

Answer 7

because p=(3(x2+1)/(2x−1))^2 gives \[((9x^4 +18x^2 +9) / (4x^2 - 4x + 1))\]

Answer 8

u havent added in -3(p + 3)

Answer 9

and how you do that

Answer 10

\[\frac{ (9x+18x ) } { (2 x-1) }\]

Answer 11

\[\frac{ 9x^2 +18x }{ 2x -1 }\times \frac{ 2x -1 }{2x -1 }\]

Answer 12

solutin pls im still confused?

Answer 13

\[= \frac{ 18 x^3+27 x^2-18 x }{4x^2 - 4x +1 }\]

Answer 14

as the 2 equations now have the same denominator, u can add the numerators together

Answer 15

\[(9x^4 +18x^2 + 9) - (18x^3+27x^2−18x)\]

Answer 16

\[ = (9x^4 -18x^3 +18x^2 - 27x^2 + 18x + 9) \]