Question

Help solve

Answer 1

\[(x-4)^{2}=49\]

Answer 2

take the square root of both sides
don't forget the \(\pm\)
add 4 to both sides

Answer 3

\(\Large \color{MidnightBlue}{\Rightarrow (a \pm b)^2 = a^2 \pm 2ab + b^2 }\)

Use this identity/formula

Answer 4

Well yes sat has a point for a simpler technique

Answer 5

i wouldn't use second method, makes life more complicated
this is a two step problem

Answer 6

do not expand the brackets

Answer 7

Yes. Just sqrt both sides lol

Answer 8

Yeah.. I am lost. What?

Answer 9

square root of 49 is....

Answer 10

\(\Large \color{MidnightBlue}{\Rightarrow x - 4 = \pm? }\)

Answer 11

I know what the square root of 49 is. I dont understand how to set the problem up.

Answer 12

\[(x−4)^2=49\]\[\sqrt{(x−4)^2}=\sqrt{49}\]\[\pm(x-4)=7\]

Answer 13

\[(x-4)^2=49\] therefore \(x-4=7\) or \(x-4=-7\)

Answer 14

This has two solutions. You take the square root of both sides.

Answer 15

if \(x-4=7\) then \(x=11\) and if \(x-4=-7\) then \(x=-3\)

Answer 16

You have two different equations. Solve for x in both of them.

Answer 17

\[(x+3)^2=36\]
\[x+3=6\] or
\[x+3=-6\]
\[x=3\] or
\[x=-9\]

Answer 18

\[(x-2)^2=10\]
\[x-2=\sqrt{10}\] or
\[x-2=-\sqrt{10}\]
\[x=2+\sqrt{10}\] or
\[x=2-\sqrt{10}\]

Answer 19

I need help on another problem like this.

Answer 20

etc
once you have the form \((x+a)^2=b\) you solve in two steps

Answer 21

If it is a perfect square on the RHS then you can take the sqrt of both sides easily.

Answer 22

go ahead and post it, can knock it out in two seconds

Answer 23

Haha sat can do anything.