So I was helping an algebra student with his homework, and this quadratic came up.
[eqn]\sqrt{2}x^2+(2\sqrt{2}-2)x-4=0[/eqn]
I can verify using the ABC method and just by plugging into the function that the roots are [math]\sqrt{2}[/math] and [math]-2[/math]. But for the life of me, I can't figure out how to show this with the quadratic formula.
[eqn]x=\frac{-(2\sqrt{2}-2) \pm \sqrt{(2\sqrt{2}-2)^2-4(\sqrt{2})(-4)}}{2\sqrt{2}}[/eqn]
Any ideas how to simplify this? I've tried every method I can think (factoring out of the square root, multiplying by conjugates, etc.).
>>8199403
dont be lazy. open up the expression in the root, then divide the numerator by the denominator. that's it.
foil on the first term under the square root
>>8199403
sorry i cannot into tex but, sqrt((2sqrt(2)-2)^2-4(sqrt(2)(-4))) = 2sqrt(2)+2, and from then on it's trivial
>>8199435
you're not going to convince me the quadratic formula doesn't work. double check your procedure, grab a blank page, and do it again
>>8199440
you know there are different ways to write things that are not obviously true at all right?
If thats what you looking for.
>>8199447
>obviously
stop being lazy and work it. no one said it's obvious from inspection.
>>8199436
>>8199440
Alright, so here is where we end up
[eqn]x=\frac{-(2\sqrt{2}-2) \pm \sqrt{2\sqrt{2}+2}}{2\sqrt{2}}[/eqn]
You can break up the fraction and simplify a little
[eqn]x=\frac{-(2\sqrt{2}-2)}{2\sqrt{2}} \pm \frac{\sqrt{2\sqrt{2}+2}}{2\sqrt{2}}=-1+\frac{1}{\sqrt{2}} \pm \sqrt{\frac{\sqrt{2}}{4}+\frac{1}{4}}[/eqn]
From here, I'm not sure how you can add these three terms together. There doesn't seem to be any way to simplify the addition under the square root term.
>>8199447
Not that guy, and I understand what you are getting at.
Correct, straightforward usage of the quadratic formula is not one of those situations. OP has (I should double-check) correctly applied the quadratic formula to yield the two roots of his polynomial. He's just getting hung up on radicals because he's young or something and doesn't "trust" the quadratic formula for whatever reason, perhaps because he is confusing himself. He does not realize that by plugging and chugging his stuff into the quadratic formula, he /has already/ "shown the solutions using the quadratic formula".
What OP really needs to do is to review how to derive the quadratic formula itself, and actually plug in that value for x both times and verify that it all cancels. These two exercises will help him start trusting his somewhat onerous solution - which he could then certainly simplify somewhat if he chose!
>>8199436
Looks like I misunderstood you. How did you get this?
[eqn]\sqrt{(2\sqrt{2}-2)^2-4(\sqrt{2})(-4)}=2\sqrt{2}+2[/eqn]
>>8199456
you're almost done. add up the terms taking common denominator, and then take root of numerator and denominator separately
>>8199466
For comparison, I'm getting
[eqn]\sqrt{8+2(-2)(2\sqrt{2})+4+16\sqrt{2}}=\sqrt{12+8\sqrt{2}}[/eqn]
>>8199480
np man glad i could help.
