Question

7rootx=x

Answer 1

i dont know the sign for taking the square root
so my problem is (7* the square root of x=x)

Answer 2

Could you show us what you've done so far please?

Answer 3

nothing. im totally lost

Answer 4

Well first let's get rid of the square root sign. To do that, wouldn't we want to square both sides?

Answer 5

So
\[(7\sqrt{x})^2=(x)^2\]

Answer 6

Yes

Answer 7

It depends. I just need to ask whether you've done complex numbers before?

Answer 8

49x^2=x^2

Answer 9

I have. I just suck at algebra

Answer 10

It seems you haven't done complex numbers so we will continue. Okay. so then I can't do it this way. We use substitution.

Answer 11

WE let \[u=\sqrt{x}\]

Answer 12

Using substitution, we get:
\[7u=u^2\]

Answer 13

@jerod530707 You there? Because once we let u=sqrtx, we can just move the 7u to the RHS [Right Hand Side]

Answer 14

im very confused

Answer 15

Okay, well let's use substitution. We let:
\[u=\sqrt{x}\]
So you started with this equation:
\[7\sqrt{x}=x^2\]
Then I made the x^2 in terms of \sqrt{x}
\[7\sqrt{x}=(\sqrt{x})^2\]
\[u^2-7u=0\]
\[u(u-7)=0\]

Answer 16

Ad then we substitute the sqrt(x) with "u". SO everytime you see \sqrt{x} you put in "u".

Answer 17

And*

Answer 18

If you're still confused at the beginning or not sure how I made x^2 in terms of sqrtx, please tell me and I will try and give you a clearer explanation.

Answer 19

I aprreciate the help Azteck, but im just not getting it

Answer 20

Well for starters, have you learnt substitution when you let something equal to a more easier pronumeral such as u or k etc?

Answer 21

if i have it was a long time ago. i have done complex though

Answer 22

\[7\sqrt x=x\\0=x-7\sqrt x\]\[\\\qquad\text{let }x=z^2\\z^2-7z=0\]

\[z=\frac{-(-7)\pm\sqrt{(-7)^2-4(1)(0)}}{2(1)}\\=\frac{7\pm\sqrt{49}}2\\=\frac72\pm\frac72\\=7,0\]
\[x=7^2,0^2\]

Answer 23

For example, let's say you were given an equation:
\[x^4+x^2-2=0\]
And you want to solve for x. Your instant thought process would to picture a quadratic instead of something like what I wrote above. TO do, we would substitute something with the letter "u".
So to make that equation become some sort of quadratic, it would be good to let:
\[u=x^2\]
That way, we can turn that into an quadratic, like so:
\[(x^2)^2+x^2-2=0\]
\[(u)^2+u-2=0\]

Answer 24

To do that*

Answer 25

nevermind. Thank you though. Ill ask my teacher

Answer 26

That's how substitution works. And then after you solve the quadratic, you reverse the substitution and make u=x^2 instead.

Answer 27

Okay, well you should ask your teacher about substitution, because if you've learnt complex numbers, you would of definitely learnt substitution before.

Answer 28

square both sides and get 49x=x^2, x=49

Answer 29

@pasta you missed a solution, x=0

Square both sides
\[\Large 7^2 \sqrt x ^2 =x ^ 2\]
After squaring both sides you get \[\Large 49 x = x^2\]Now subtract 49x from both sides\[\Large 0 = x^2 - 49x\] and factor out an x \[\Large 0 = x(x-49)\]
which means either x=0 or x=49, and then you just need to check both solutions are valid in:\[\Large 7 \sqrt x = x\]

Answer 30

i got u.thankx