Question

2x^2−(2√2+1)x+√2=0
solve using the quadratic formula

Answer 1

@jim_thompson5910

Answer 2

@zepdrix

Answer 3

lol this equation again huh? :)

Answer 4

yepp

Answer 5

i started it but i can't simplify it.

Answer 6

\[\Large\rm x=\frac{-\color{orangered}{b}\pm\sqrt{\color{orangered}{b}^2-4\color{royalblue}{a}\color{#CC0033}{c}}}{2\color{royalblue}{a}}\]
And we have:\[\Large\rm \color{royalblue}{2}x^2+\color{orangered}{-(2\sqrt2+1)}x+\color{#CC0033}{\sqrt2}=0\]

Answer 7

yep i did plug in all the numbers into the formula

Answer 8

i just get so many radicals in the sqaureroot

Answer 9

\[\Large\rm x=\frac{\color{orangered}{2\sqrt2+1}\pm\sqrt{\color{orangered}{(2\sqrt2+1)}^2-4\color{royalblue}{(2)}\color{#CC0033}{(\sqrt2)}}}{2\color{royalblue}{(2)}}\]Having trouble simplifying it or something? :o

Answer 10

Hmmm

Answer 11

\[\Large\rm x=\frac{2\sqrt2+1\pm\sqrt{8+4\sqrt2+1-8\sqrt2}}{4}\]Oooo I thought those sqrt(2) would cancel out under the root :(
Hmmm

Answer 12

exactly thats where i got stuck

Answer 13

\[\Large\rm x=\frac{2\sqrt2+1\pm\sqrt{9-4\sqrt2}}{4}\]

Answer 14

Wolfram is telling me that \(\Large\rm \sqrt{9-4\sqrt2}\) is equivalent to \(\Large\rm 2\sqrt{2}-1\).

Hmm I'm trying to figure out how we get there...

Answer 15

oh

Answer 16

Hmm there is a method that will work for `Denesting` square roots.
But it's a little weird :o

Why didn't you use the grouping method? No bueno?
That works out much much easier.

Answer 17

well we're required to use grouping, quadratic formula, and completing the square, and graphing

Answer 18

Ok I'm reading a website that explains denesting square roots.

I'll post the method, it doesn't seem too bad.

Answer 19

Try not to confuse these letters with our quadratic formula letters :)
So we have something of this form:
\[\Large\rm \sqrt{9-4\sqrt2}=\sqrt{a-b\sqrt{c}}\]

We first calculate: \(\Large\rm a^2-b^2\cdot c\)
If this value turns out to be a perfect square, then we are able to simplify our expression.

So for our problem we get: \(\Large\rm 9^2-4^2\cdot 2=49\)
Which is \(\Large\rm 7^2\) and we'll call this \(\Large\rm q^2\).
So \(\Large\rm 7=q\).

So we're allowed to rewrite our expression like this:
\[\Large\rm \sqrt{a-b\sqrt{c}}=\sqrt{\frac{a+q}{2}}+\sqrt{\frac{a-q}{2}}\]So for our problem:\[\Large\rm \sqrt{9-4\sqrt{2}} =\sqrt{\frac{9+7}{2}}+\sqrt{\frac{9-7}{2}}\]

Answer 20

Which simplifies a bit further, yes? :o
\[\Large\rm =\sqrt{8}+\sqrt{1}\]\[\Large\rm =2\sqrt{2}+1\]

Answer 21

but how do you get 1/2 and square root 2

Answer 22

So that only takes care of this orange part:\[\Large\rm x=\frac{2\sqrt2+1\pm\color{orangered}{\sqrt{9-4\sqrt2}}}{4}\]
Which becomes:\[\Large\rm x=\frac{2\sqrt2+1\pm\color{orangered}{(2\sqrt2+1)}}{4}\]

Answer 23

And now you need to look at the plus/minus separately to get your two solutions.\[\Large\rm x=\frac{2\sqrt2+1\color{royalblue}{\pm}(2\sqrt2+1)}{4}\]Turns into two equations:\[\Large\rm x=\frac{2\sqrt2+1\color{royalblue}{+}(2\sqrt2+1)}{4}\]and\[\Large\rm x=\frac{2\sqrt2+1\color{royalblue}{-}(2\sqrt2+1)}{4}\]

Answer 24

AHhhh I did the wrong simplification for the root :(
It can also be written as:
\(\Large\rm =\sqrt{\frac{a+q}{2}}-\sqrt{\frac{a-q}{2}}\)
That's the one we wanted ^
With the minus between the terms.

Lemme fix that in our setup.

Answer 25

Ok so these are our two equations:
\[\Large\rm x=\frac{2\sqrt2+1\color{royalblue}{+}(2\sqrt2-1)}{4}\]and\[\Large\rm x=\frac{2\sqrt2+1\color{royalblue}{-}(2\sqrt2-1)}{4}\]

Answer 26

Do you see how the `1`'s will cancel out in the `plus` equation,

and how the `2sqrt(2)`'s will cancel out in the `minus` equation?

Answer 27

Quadratic Formula REALLY sucks for these types of roots :O( lol

Answer 28

yelp could you also help solve it completing the square and graping(wolfram)/

Answer 29

What type of class is this for?
If it's for a Calculus class, we can get a more accurate graph by finding the max/min values.

Otherwise if it's only for Algebra or something, we can graph the turning points and then use a graphing calculator or something to rough in the rest.

Answer 30

Completing the square?
Oh boy, that's going to be awful lolol

Answer 31

\[\Large\rm 2x^2-(2\sqrt2+1)x+\sqrt2=0\]Subtract sqrt2 to the other side, then divide both sides by 2,\[\Large\rm x^2-\left(\frac{2\sqrt2+1}{2}\right)x\qquad\qquad\qquad=-\frac{\sqrt{2}}{2}\]To complete the square, we take half of our b coefficient, and square it.
That's the value we want to add to each side.\[\Large\rm \left(\frac{b}{2}\right)^2=\left(-\frac{2\sqrt2+1}{4}\right)^2=\left(\frac{2\sqrt2+1}{4}\right)^2\]So we add that to each side,\[\large\rm x^2-\left(\frac{2\sqrt2+1}{2}\right)x+\left(\frac{2\sqrt2+1}{4}\right)^2=-\frac{\sqrt{2}}{2}+\left(\frac{2\sqrt2+1}{4}\right)^2\]Our perfect square will condense down to \(\Large\rm \left(x+\frac{b}{2}\right)^2\),
giving us,\[\large\rm \left(x-\frac{2\sqrt2+1}{4}\right)^2=-\frac{\sqrt{2}}{2}+\left(\frac{2\sqrt2+1}{4}\right)^2\]

Answer 32

And then do some square root business,\[\large\rm x-\frac{2\sqrt2+1}{4}=\pm\sqrt{-\frac{\sqrt{2}}{2}+\left(\frac{2\sqrt2+1}{4}\right)^2}\]

Answer 33

And then some more steps and some things....
It'll probably work out similar to the Quadratic Formula method from here.

I dunno, I'm too tired :d
Time to call it a night I think >.<