Quadratic Equations (that can't be Factorised)

Just be careful with your notation. You used b^2-4ac in your attempt at a solution section, but then switched to (b2 - 4ac) in your summary. It's important to use consistent notation and make sure all symbols are properly defined. Other than that, great job summarizing the conversation! In summary, the question asks to prove that p2 - 3(p+3) \geq 0 given that x is real and p = \frac{3(x^{2}+1)}{2x-1}. The attempt at a solution correctly simplifies the equation and shows that if b^2-4ac \geq 0, then x has real solutions. Using the ABC formula
  • #1
mcsun
4
0
Homework Statement

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.

Homework 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.

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?
 
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  • #2
Everything you have done so far is correct. So let's 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 let's take a real number, let's 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:
  • #3
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
 
  • #4
You're welcome.
 

1. What is a quadratic equation?

A quadratic equation is a polynomial equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is a variable. It can also be written as y = ax^2 + bx + c. The highest power in the equation is 2, hence the name quadratic.

2. How do you know if a quadratic equation can't be factorised?

A quadratic equation can't be factorised if its discriminant (b^2 - 4ac) is less than 0. This means that the equation has no real solutions, and hence, can't be factored into two linear expressions.

3. What methods can be used to solve a quadratic equation that can't be factorised?

There are two main methods to solve a quadratic equation that can't be factorised: using the quadratic formula and completing the square. The quadratic formula is (-b ± √(b^2 - 4ac)) / 2a, and completing the square involves manipulating the equation to make it in the form of (x + p)^2 = q.

4. Is there a way to determine the number of solutions for a quadratic equation that can't be factorised?

Yes, the number of solutions for a quadratic equation that can't be factorised can be determined by looking at the discriminant. If the discriminant is greater than 0, there are two real solutions. If it is equal to 0, there is one real solution. And if it is less than 0, there are no real solutions.

5. How can quadratic equations be used in real-life situations?

Quadratic equations can be used to model various real-life situations, such as calculating the maximum or minimum value of a function, determining the trajectory of a projectile, or finding the optimal solution in business and economics problems. They are also used in fields like engineering, physics, and computer science.

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