- #1

- 4

- 0

**Homework Statement**

If

*x*is real and

*p*= [tex]\frac{3(x^{2}+1)}{2x-1}[/tex], prove that p

^{2}- 3(p+3) [tex]\geq[/tex] 0.

## Homework Equations

ax

^{2}+ bx + c = 0

x = [tex]\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}[/tex]

Nature of the Roots of a Quadratic Equation:

If b

^{2}- 4ac is positive, quadratic equation has two real & distinct roots.

If b

^{2}- 4ac is zero, quadratic equation has repeated roots or equal roots.

If b

^{2}- 4ac is negative, quadratic equation has no real roots.

## The Attempt at a Solution

p = [tex]\frac{3(x^{2}+1)}{2x-1}[/tex]

p(2x - 1) = 3(x

^{2}+ 1)

3x

^{2}- 2px + (p + 3) = 0

Now to determine the nature of the root:

b

^{2}- 4ac = (-2p)

^{2}- 4(3)(p + 3)

b

^{2}- 4ac = 4[p

^{2}- 3(p + 3)]

After this stage, i don't know how to continue on to prove that p

^{2}- 3(p+3) [tex]\geq[/tex] 0. Or is my above strategy wrong?