1. ## Quadratic Equation general question

As we know the standard form of a quadratic equation is $ax^2 + bx + c = 0$

But when B is equal to 0 in the above it looks like this $ax^2 =c$. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

Take this for example

$x^2 = 49$ Obviously you could just do the $\sqrt49$ to get +- 7. But What if you wanted to put it in general quadratic form?

If I wanted to put it general quadratic form would this be correct?
$x^2 - 49 = 0$
$(x + 7)(x - 7) = 0$

Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

So in conclusion has my reasoning and calculations been correct up to this point?

Many thanks!

2. Originally Posted by cmf0106
As we know the standard form of a quadratic equation is $ax^2 + bx + c = 0$

But when B is equal to 0 in the above it looks like this $ax^2 =c$. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

Take this for example

$x^2 = 49$ Obviously you could just do the $\sqrt49$ to get +- 7. But What if you wanted to put it in general quadratic form?

If I wanted to put it general quadratic form would this be correct?
$x^2 - 49 = 0$
$(x + 7)(x - 7) = 0$

Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

So in conclusion has my reasoning and calculations been correct up to this point?

Many thanks!
Um, yeah, pretty much

3. Originally Posted by cmf0106
As we know the standard form of a quadratic equation is $ax^2 + bx + c = 0$

But when B is equal to 0 in the above it looks like this $ax^2 =c$. However to get the c in the preceding to look like a standard quadratic equation you must add its inverse to the other side.

Take this for example

$x^2 = 49$ Obviously you could just do the $\sqrt49$ to get +- 7. But What if you wanted to put it in general quadratic form?

If I wanted to put it general quadratic form would this be correct?
$x^2 - 49 = 0$
$(x + 7)(x - 7) = 0$

Then just use the multiplicative property of zero to solve both equations and arrive at +- 7?

So in conclusion has my reasoning and calculations been correct up to this point?

Many thanks!
Yes, up to now your conclusion has been correct, but even more generally

in $ax^2+b+c=0$ if we let $b=0$

then by the quadratic formula the solution is

$\frac{0\pm\sqrt{0-4(a)(c)}}{2a}$

Which shows the fact that for a case where b=0 one of the two must have a negative coefficient otherwise ti would yield an imaginary result