Quadratics

[Lemon Of Life]

Sorry that this isn't a subject of much interest or potential discussion, but I have a maths exam tomorrow, and would greatly appreciate it if you put your great minds to good use.

When faced with a quadratic equation such as h^2+7h-72=0 and asked to show that it equals zero, do you simply solve it or must another route of action be taken?

Any help = much appreciated.

EDIT: Also, when dealing with simultaneous equations, what do you do if one of them has squares, such as in:

y=2x+3

x^2+y^2=2

 

[Redingold]

You can use the quadratic formula.

 

[Lemon Of Life]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

 

[Good morning blues]

I don't see what you could do other than solve the problem.

 

[Zukhramm]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

You can't really, as it only equals zero for two particular h, then only thing you can do is find for which h the equation does equal zero.

 

[Trivun]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

It doesn't, you simply input the x-value back into the original equations to prove it. It's known as 'testing the formula'.

As far as squared functions go, you simply square both sides of the first function. So in your example:

y=2x+3
x^2+y^2=2

You take y^2=(2x+3)^2. Then you use that in the second function to get a single quadratic equation in terms of x. Then when you find x using the quadratic formula, or expanding brackets, then you can use that to find the values of y.

 

[Lemon Of Life]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

It doesn't, you simply input the x-value back into the original equations to prove it. It's known as 'testing the formula'.

As far as squared functions go, you simply square both sides of the first function. So in your example:

y=2x+3
x^2+y^2=2

You take y^2=(2x+3)^2. Then you use that in the second function to get a single quadratic equation in terms of x. Then when you find x using the quadratic formula, or expanding brackets, then you can use that to find the values of y.

Cheers, big help.

 

[Zukhramm]

y=2x+3

x^2+y^2=2

I think there's a couple of way to to this, but what I'd do is rearange one of the to find one with "y = the rest", which the ones you've provided already have. Since y is (2x+3), just replace the y in the other equation with that and go from there.

 

[Nimbus]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

It doesn't, you simply input the x-value back into the original equations to prove it. It's known as 'testing the formula'.

As far as squared functions go, you simply square both sides of the first function. So in your example:

y=2x+3
x^2+y^2=2

You take y^2=(2x+3)^2. Then you use that in the second function to get a single quadratic equation in terms of x. Then when you find x using the quadratic formula, or expanding brackets, then you can use that to find the values of y.

Would it not be simpler to take the Y in terms of X in the first equation and substitute it into the quadratic equation?

 

[Xyliss]

Sorry that this isn't a subject of much interest or potential discussion, but I have a maths exam tomorrow, and would greatly appreciate it if you put your great minds to good use.

When faced with a quadratic equation such as h^2+7h-72=0 and asked to show that it equals zero, do you simply solve it or must another route of action be taken?

Any help = much appreciated.

EDIT: Also, when dealing with simultaneous equations, what do you do if one of them has squares, such as in:

y=2x+3

x^2+y^2=2

Yes it will give you the two solutions where it equals zero.

And as for the simultaneous equations just sub the y= equation into the second one then square it (make sure you square the whole thing so in your example it would be (2x+3)(2x+3)=4x^2+12x+9 for y^2
Then you just solve for x and sub back in to find y

 

[Redingold]

Wait, wait, wait. I just read your post again. You want to know how to prove that h[sup]2[/sup]+7h-72=0, equals zero? It says so right there in the equation, that bit with the equals sign.

 

[Xyliss]

Wait, wait, wait. I just read your post again. You want to know how to prove that h[sup]2[/sup]+7h-72=0, equals zero? It says so right there in the equation, that bit with the equals sign.

That ain't a proof if they ask it. It means he needs to show it...without just highlighting the equals sign

 

[Zukhramm]

Would it not be simpler to take the Y in terms of X in the first equation and substitute it into the quadratic equation?

That's essentially what he's doing.

 

[Redingold]

Wait, wait, wait. I just read your post again. You want to know how to prove that h[sup]2[/sup]+7h-72=0, equals zero? It says so right there in the equation, that bit with the equals sign.

That ain't a proof if they ask it. It means he needs to show it...without just highlighting the equals sign

It kind of is a proof. If it says that it equals zero, then it can't not equal zero. I don't see how saying that it equals zero is not proof that it equals zero.

 

[thethingthatlurks]

Normally the directions "show that F(x) equals zero) means that F(x) = 0 for any value of x. I guess there's some language barrier here, because that isn't possible for a quadratic. Just bring all the terms over to one side (as you've done) and solve it using the quadratic equation. And be mindful of the signs.
Cheers and good luck!

 

[Trivun]

You can use the quadratic formula.

I know how to solve it, but would that prove that it equals zero?

It doesn't, you simply input the x-value back into the original equations to prove it. It's known as 'testing the formula'.

As far as squared functions go, you simply square both sides of the first function. So in your example:

y=2x+3
x^2+y^2=2

You take y^2=(2x+3)^2. Then you use that in the second function to get a single quadratic equation in terms of x. Then when you find x using the quadratic formula, or expanding brackets, then you can use that to find the values of y.

Would it not be simpler to take the Y in terms of X in the first equation and substitute it into the quadratic equation?

Ummmmmm, that's exactly what I've done...

 

[Lemon Of Life]

Wait, wait, wait. I just read your post again. You want to know how to prove that h[sup]2[/sup]+7h-72=0, equals zero? It says so right there in the equation, that bit with the equals sign.

That ain't a proof if they ask it. It means he needs to show it...without just highlighting the equals sign

It kind of is a proof. If it says that it equals zero, then it can't not equal zero. I don't see how saying that it equals zero is not proof that it equals zero.

That's all very good and clever, but I've got a question here on a past paper telling me to prove it, and that won't quite cut it.

 

[bhenders13]

for y=2x+3
x^2+y^2=2, the solution set would either be (-7/5, 1/5) or (-1, 1). For h^2+7h-72=0, than h would either equal 5.678779875343 or -12.678779875343. I got h by using the quadratic formula. the reason it is such a huge decimal is because in the calculator, that is what rad 337 is. If you want work i can provide it.

 

[Xyliss]

Wait, wait, wait. I just read your post again. You want to know how to prove that h[sup]2[/sup]+7h-72=0, equals zero? It says so right there in the equation, that bit with the equals sign.

That ain't a proof if they ask it. It means he needs to show it...without just highlighting the equals sign

It kind of is a proof. If it says that it equals zero, then it can't not equal zero. I don't see how saying that it equals zero is not proof that it equals zero.

Not if they're asking you to show it. I could right 1=2 and you could call that a proof, but it isn't you need something to back it up...which is what they're asking (although to be honest what you show them isn't a proof either...it just proves it could work)

 

[Eclectic Dreck]

I don't see what you could do other than solve the problem.

Then you would prove that your two solutions are at 0 and again at zero. y=x^2 is an example of this very thing happening. The quadratic formula also helps identify non-real solutions. In any example of a quadratic equation, if it does not cross a given axes your solution is imaginary.