You are Here: Home >< Maths

Announcements Posted on
Four hours left to win £100 of Amazon vouchers!! Don't miss out! Take our short survey to enter 24-10-2016
1. While going through a physics problem, I found following mathematical property being used.

If a > 0 and b2 - 4ac < 0, then ax2 + bx + c > 0 for all x.

Can anyone explain, why the property holds true?
2. (Original post by tangotangopapa2)
While going through a physics problem, I found following mathematical property being used.

If a > 0 and b2 - 4ac < 0, then ax2 + bx + c > 0 for all x.

Can anyone explain, why the property holds true?
Are you familiar with the formula for the roots of a quadratic equation?
3. (Original post by ghostwalker)
Are you familiar with the formula for the roots of a quadratic equation?
Yes. I am.
4. is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if then the equation has complex roots (non real), if then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if then the equation has 2 distinct real roots.
The general solution to the quadratic you gave is . So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the x-axis or not.
If then the curve y does not cross the x-axis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis.
For example if is , what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
5. (Original post by tangotangopapa2)
Yes. I am.
Good, that saves a lot of LaTex.

b2 - 4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the x-axis. So, it is always positive or always negative.

If, in addition, a>0, then as x goes off to infinity, so ax2 + bx + c goes off to infinity, i.e. it's positive from some point onwards.

Putting the two together, since it must be positive at some point, it is always positive.
6. Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.

b2 - 4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.

It's shown as part of the quadratic formula

where it's evident that it cannot be solved if the discriminant is less than zero.

It's actually derived by completing the square. For the equation ax2 + bx + c = 0:

x^2 + (b/a)x + c/a = 0

(x + b/2a)2 - (b/2a)2 + c/a = 0

(x + b/2a)2 = (b/2a)2 - c/a

As the LHS is a square:

(b/2a)2 - c/a ≥ 0

(b2/4a2) - c/a 0

(b2/4a2) - 4ac/4a2 0

(b2 - 4ac)/4a2 0

And therefore, as 4a2 is always positive, b2 - 4ac 0 for there to be real solutions of x.
7. (Original post by B_9710)
is called the disriminant of the quadratic equation, it tells us of the nature of the roots, if then the equation has complex roots (non real), if then the equation has a repeated root (one root basically where the curve is tangent to the x-axis) and if then the equation has 2 distinct real roots.
The general solution to the quadratic you gave is . So if the expression inside the square root is negative then it cannot possibly be a real number. Basically the discriminant tells us the curve crosses the x-axis or not.
If then the curve y does not cross the x-axis, if a>0 this means that the curve lies completely above the x axis. If a<0 it means that the curve is always under the x axis.
For example if is , what would the value of x be? Certainly no real number, so it cannot cross the x axis right.
Thank you so much. Got it.
8. (Original post by ghostwalker)
Good, that saves a lot of LaTex.

b2 - 4ac < 0 implies tha the quadratic has no real roots, i.e. it does not cross the x-axis. So, it is always positive or always negative.

If, in addition, a>0, then as x goes off to infinity, so ax2 + bx + c goes off to infinity, i.e. it's positive from some point onwards.

Putting the two together, since it must be positive at some point, it is always positive.
Thanks alot.
9. (Original post by Alexion)
Well, first, that's incorrect - if b2 - 4ac < 0 then there's no real solution of x.

b2 - 4ac is known as the discriminant of the quadratic. If < 0, there's real solutions, if = 0, there's two identical solutions (one, for all intents and purposes), and if > 0, then there's two different real solutions of the quadratic.

It's shown as part of the quadratic formula

where it's evident that it cannot be solved if the discriminant is less than zero.

It's actually derived by completing the square. For the equation ax2 + bx + c = 0:

x^2 + (b/a)x + c/a = 0

(x + b/2a)2 - (b/2a)2 + c/a = 0

(x + b/2a)2 = (b/2a)2 - c/a

As the LHS is a square:

(b/2a)2 - c/a ≥ 0

(b2/4a2) - c/a 0

(b2/4a2) - 4ac/4a2 0

(b2 - 4ac)/4a2 0

And therefore, as 4a2 is always positive, b2 - 4ac 0 for there to be real solutions of x.
Thank you very much.
10. (Original post by Alexion)
if = 0, there's two identical solutions (one, for all intents and purposes)

???

It's simple - there's just one. Not 2 identical ones
11. There is also the following formula that no-one uses.
12. (Original post by tangotangopapa2)
Thank you very much.
Is the converse also true?
i.e if the equation is positive for all values of x then, a>0 and discriminant<0?
Yes, one implies the other.
13. (Original post by Student403)
???

It's simple - there's just one. Not 2 identical ones
All quadratic equations have two solutions as a fundamental rule - so both solutions are the same
14. (Original post by Ano123)
There is also the following formula that no-one uses.
Cool. Took the ratio of two formulas and ans is 1. Verified.
I'm gonna use it where ever possible . :d
15. (Original post by tangotangopapa2)
Cool. Took the ratio of two formulas and ans is 1. Verified.
I'm gonna use it where ever possible . :d
They are just really the same formula, just rationalise the denominator to get to the more conventional one. Advantage of this one though is that it gives you a solution even if a=0.
16. (Original post by Ano123)
They are just really the same formula, just rationalise the denominator to get to the more conventional one. Advantage of this one though is that it gives you a solution even if a=0.
Ah. It falsely gives you two roots for linear equation. Though the second root is undefined, does it not violate one of the fundamental theorems of algebra? A nth degree equation has n-roots, no more.

Anyway thanks.
17. (Original post by tangotangopapa2)
Ah. It falsely gives you two roots for linear equation. Though the second root is undefined, does it not violate one of the fundamental theorems of algebra? A nth degree equation has n-roots, no more.

Anyway thanks.
You would have to take the positive root of the b^2 because otherwise you would get 0 on the denominator. It gives x=-c/b which of course is the solution to bx+c=0.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
3. this can't be left blank

6 characters or longer with both numbers and letters is safer

4. this can't be left empty
1. Oops, you need to agree to our Ts&Cs to register

Updated: July 5, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### Who is getting a uni offer this half term?

Find out which unis are hot off the mark here

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams