Can I get some more help? I'm confused. √2x-7=4
\[\huge{\color{purple}{\textbf{W}} \color{orange}{\cal{E}} \color{green}{\mathbb{L}} \color{blue}{\mathsf{C}} \color{maroon}{\rm{O}} \color{red}{\tt{M}} \color{gold}{\tt{E}} \space \color{orchid}{\mathbf{T}} \color{Navy}{\mathsf{O}} \space \color{OrangeRed}{\boldsymbol{O}} \color{Olive}{\mathbf{P}} \color{Lime}{\textbf{E}} \color{DarkOrchid}{\mathsf{N}} \color{Tan}{\mathtt{S}} \color{magenta}{\mathbb{T}} \color{goldenrod}{\mathsf{U}} \color{ForestGreen}{\textbf{D}} \color{Salmon}{\mathsf{Y}} \ddot \smile }\] @Taylorcheer98 what do you think we do first
The 2x-7?
Wait, is it \(\sqrt{2x-7} = 4\) or \(\sqrt{2x} - 7 = 4\) ?
^^
It's the first one. What is the difference between the two?
A huge difference.
It could get you a wrong answer. Always good to clarify :)
Could you tell me the difference between the two
first we square both sides of the equation.
Well, we know that \(\sqrt{2x-7}=4\) is the same as \(2x-7 = 4^2\) , right?
you got this @sammixboo ?
If you want to continue I'll leave :o
=3
its ok
Yes
Well @Taylorcheer98 the first one is easier to solve, the second takes more steps. And you will get 2 different answers. Because notice how the 7 or could not be under the squareroot?
light.
what he said
Yes
OK so we have \(2x-7 = 4^2\) Let's do the exponent, \(4^2\) first. Do you know what that is equal to? \(4^2\rightarrow 4\times 4\)
16?
Right so \(2x - 7 = 16\) Do you know what to do next? Solve this like a regular equation
Ok, let me solve it
OK tell us what you get, so we can check your answer x3
23/2?
Correct!!
Thank you I understand it better'!
=3 if you need anymore help tag me @sammixboo
I definitely will.
Okie dokie! Welcome to OpenStudy, once again! If you have any concerns on how to use the site, Private Message me! I will reply ASAP
Thank you so much! I appreciate that.
:)
Join our real-time social learning platform and learn together with your friends!