Question

square root of


x= √2x-33

Answer 1

Do you mean to solve for x in:
\[x=\sqrt{2x-33}\]

Answer 2

\[x^2=2x-33\]
\[x^2-2x+33=0\]

Answer 3

do you mean \[\sqrt{x}=\sqrt{2x-33}\]?

Answer 4

I mean Taking the Square root of x²=2x+33
which I put into x= √2x-33


and now I'm stuck.

Answer 5

\[[2pmsqrt{-128}]/2 complex roots\]

Answer 6

x^2-2x+33=0 You will need to use the quadratic formula to solve this

Answer 7

x^2-2x+33=0 gives complex roots.
Do you work with complex roots, Mandy?

Answer 8

\[x^{2}-2x+33=(x-1)^{2} +31>=31 >0 \] so it dont have real roots

Answer 9

"I mean Taking the Square root of x²=2x+33 which I put into x= √2x-33"
We don't really understand by your statement above. You can help to clarify by putting parentheses.
For example, "square root of x^2" or "square root of (x^2=2x+33)"?
I do not understand either interpretation.

Answer 10

\[x ^{2}=2x-33 => x=\sqrt{(2x-33)}=>x=(2x-33)^{1/2}\]

Answer 11

to solve for the roots of x
\[x=\sqrt{2x-33} => x ^{2}-2x+33=0 (Zed )\]
Using,\[x=(-b \pm \sqrt{b ^{2}-4ac})/2a\]
\[a=1, b=-2, c=33\]
x=\[(-(-2)\pm \sqrt{(-2)^2-4(1)*(33)})/2(1)\]
x=\[(2\pm \sqrt{-128})/2\]

Answer 12

\[x=(2+j8\sqrt{2}), (2-j8\sqrt{2})/2 => x=1+j4\sqrt{2}, 1-j4\sqrt{2}\]