# Quadratic Equation Help

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While going through a physics problem, I found following mathematical property being used.

If a > 0 and b

Can anyone explain, why the property holds true?

Thanks in advance.

If a > 0 and b

^{2}- 4ac < 0, then ax^{2}+ bx + c > 0 for all x.Can anyone explain, why the property holds true?

Thanks in advance.

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#2

(Original post by

While going through a physics problem, I found following mathematical property being used.

If a > 0 and b

Can anyone explain, why the property holds true?

Thanks in advance.

**tangotangopapa2**)While going through a physics problem, I found following mathematical property being used.

If a > 0 and b

^{2}- 4ac < 0, then ax^{2}+ bx + c > 0 for all x.Can anyone explain, why the property holds true?

Thanks in advance.

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(Original post by

Are you familiar with the formula for the roots of a quadratic equation?

**ghostwalker**)Are you familiar with the formula for the roots of a quadratic equation?

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#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.

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.

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#5

(Original post by

Yes. I am.

**tangotangopapa2**)Yes. I am.

b

^{2}- 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 ax

^{2}+ 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.

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#6

Well, first, that's incorrect - if

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

As the LHS is a square:

And therefore, as 4a

**b**^{2}**- 4ac < 0**then there's no real solution of x.**b**^{2}**- 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

**ax**^{2}**+ 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****(b**^{2}**/4a**^{2}**) -****c/a****≥****0****(b**^{2}**/4a**^{2}**) -****4ac/4a**^{2}**≥****0****(b**^{2}**- 4ac)/4a**^{2}**≥****0**And therefore, as 4a

^{2}is always positive,**b**^{2}**- 4ac****≥****0**for there to be real solutions of x.
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(Original post by

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.

**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.

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(Original post by

Good, that saves a lot of LaTex.

b

If, in addition, a>0, then as x goes off to infinity, so ax

Putting the two together, since it must be positive at some point, it is always positive.

**ghostwalker**)Good, that saves a lot of LaTex.

b

^{2}- 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 ax

^{2}+ 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.

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(Original post by

Well, first, that's incorrect - if

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

As the LHS is a square:

And therefore, as 4a

**Alexion**)Well, first, that's incorrect - if

**b**^{2}**- 4ac < 0**then there's no real solution of x.**b**^{2}**- 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

**ax**^{2}**+ 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****(b**^{2}**/4a**^{2}**) -****c/a****≥****0****(b**^{2}**/4a**^{2}**) -****4ac/4a**^{2}**≥****0****(b**^{2}**- 4ac)/4a**^{2}**≥****0**And therefore, as 4a

^{2}is always positive,**b**^{2}**- 4ac****≥****0**for there to be real solutions of x.But question does not ask for equality. It asks for inequality.

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#10

(Original post by

if = 0,

**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

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#12

(Original post by

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?

**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?

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#13

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I'm gonna use it where ever possible . :d

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#15

(Original post by

Cool. Took the ratio of two formulas and ans is 1. Verified.

I'm gonna use it where ever possible . :d

**tangotangopapa2**)Cool. Took the ratio of two formulas and ans is 1. Verified.

I'm gonna use it where ever possible . :d

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(Original post by

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.

**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.

Anyway thanks.

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#17

(Original post by

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.

**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.

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