Question

help with problems medal given. have the answers just solve the problem
1. square root x-8=2
2. square root s+13= square root 29-s
3. x= square root 2x +24
5. y= square root 7x-4

Answer 1

in num 1
just squared both sides to remove square root
(square root x-8)^2=2^2
x-8=4
x=4+8

Answer 2

just the same way in all num 2
square root s+13= square root 29-s
squared both sides
then if we squared square roots it will cancel
so
s+13=29-s
s+s=29-13
2s=16
divide by 2 both sides
s=

Answer 3

\[x=\sqrt{2x+24}\] square and get
\[x^2=2x+24\] or
\[x^2-2x-24=0\] solve by factoring then check the result

Answer 4

you get
\[(x+4)(x-6)=0\] making \(x=-4\) or \(x=6\)
substitute back in the original equation, you will see that \(6\) works because
\[6=\sqrt{2\times 6+24}\] is true, but \(-4\) does not work

Answer 5

thankk you so much @satellite73 you really helped with these simple questions which other ppl couldnt do!