SOLUTION: Please help me solve by taking roots of of both sides: y^2=16 (x-3)^2=25

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Question 145132: Please help me solve by taking roots of of both sides:
y^2=16
(x-3)^2=25
Found 2 solutions by Earlsdon, solver91311:
Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve:
y%5E2+=+16 Take the square root of both sides.
sqrt%28y%5E2%29+=+sqrt%2816%29
y+=+4 or y+=+-4
%28x-3%29%5E2+=+25 Take the square root of both sides.
sqrt%28%28x-3%29%5E2%29+=+sqrt%2825%29
x-3+=+5 or x-3+=+-5 Add 3 to both sides in each case.
x+=+3%2B5 or x+=+3-5
x+=+8 or x+=+-2

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
To take the square root of a number, you need to answer the question: What other number when multiplied by itself gives you the original number.

So what number multiplied by itself gives you y%5E2? Well, we know that y%5E2=y%2Ay, so there is part of the answer to that part, namely: sqrt%28y%5E2%29=y. You also have to consider that y%5E2=%28-y%29%28-y%29 also. So sqrt%28y%5E2%29=y OR sqrt%28y%5E2%29=-y.

Apply this same logic to the number 16. What number, multiplied by itself, makes 16? Remember that 4%2A4=16 and -4%2A-4=16.

So now we have: y=4 or y=-4 (You can also say -y=4 or -y=-4, but those are just equivalent expressions to the ones already given.)

Do the other problem the same way. You'll get x-3 equal to plus or minus something, so you will need to solve each of those resulting equations by adding 3 to both sides.