# Quadratic Equations (that can't be Factorised)

Homework Statement

If x is real and p = $$\frac{3(x^{2}+1)}{2x-1}$$, prove that p2 - 3(p+3) $$\geq$$ 0.

## Homework Equations

ax2 + bx + c = 0

x = $$\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$

Nature of the Roots of a Quadratic Equation:

If b2 - 4ac is positive, quadratic equation has two real & distinct roots.

If b2 - 4ac is zero, quadratic equation has repeated roots or equal roots.

If b2 - 4ac is negative, quadratic equation has no real roots.

## The Attempt at a Solution

p = $$\frac{3(x^{2}+1)}{2x-1}$$

p(2x - 1) = 3(x2 + 1)

3x2 - 2px + (p + 3) = 0

Now to determine the nature of the root:

b2 - 4ac = (-2p)2 - 4(3)(p + 3)

b2 - 4ac = 4[p2 - 3(p + 3)]

After this stage, i don't know how to continue on to prove that p2 - 3(p+3) $$\geq$$ 0. Or is my above strategy wrong?

Cyosis
Homework Helper
Everything you have done so far is correct. So lets continue from your last step.

$$b^2 - 4ac = 4[p^2 - 3(p + 3)]$$

You stated in the relevant equation section that if $b^2-4ac \geq 0$ then x has real solutions. So lets take a real number, lets say $n \geq 0$ so that $b^2-4ac=n$. Do you understand that for this n x always has a real root? Now solve $n = 4[p^2 - 3(p + 3)]$ for $p^2-3(p+3)$. What can you conclude now?

Alternatively calculate x by using the ABC formula so you get the following equation for x:

$$x= \frac{2p \pm \sqrt{4p^2-4*3(p+3)}}{6}$$. Do you now see why it follows that $p^2-3(p+3)\geq 0$?

Last edited:
Ohhhhh... now i get it!

Since the question has already informed us that x is real for the following formula:

x = $$\frac{-b + \sqrt{b^{2} - 4ac}}{2a}$$

therefore,

(b2 - 4ac) can either be equal or greater than zero, so as to produce real roots for this quadratic equation.

Thanks Cyosis for the guidance!

mcsun

Cyosis
Homework Helper
You're welcome.