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Quadratic Equations (that can't be Factorised)

  • Thread starter mcsun
  • Start date
The problem statement, all variables and given/known data

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

2. Relevant equations

ax2 + bx + c = 0

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

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.

3. The attempt at a solution

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

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) [tex]\geq[/tex] 0. Or is my above strategy wrong?
 

Cyosis

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

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

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


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

[tex] x= \frac{2p \pm \sqrt{4p^2-4*3(p+3)}}{6}[/tex]. Do you now see why it follows that [itex]p^2-3(p+3)\geq 0[/itex]?
 
Last edited:
Ohhhhh... now i get it! :smile:

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

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

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
1,495
0
You're welcome.
 

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