Quadratic Equations (that can't be Factorised)

[mcsun]

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]

Everything you have done so far is correct. So let's 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 let's take a real number, let's 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?

 

[mcsun]

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]

You're welcome.