Thread: couple math 95 problems

[SIZE=3]& # 8 7 3 0 ;
x-2(√(x-3) = 3
so I substract the x from both sides but then what? do I square both sides or do I divide by 2 and make the other side a fraction?

√(5x+1) = x+1
I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?

also why does √x-3 not equal x+9 ? isn't -3*-3 = +9?

dang is there a way to make the square root symbol extend across the numbers?

Originally Posted by harnar

[SIZE=3]& # 8 7 3 0 ;
x-2(√(x-3) = 3
so I substract the x from both sides but then what? do I square both sides or do I divide by 2 and make the other side a fraction?

√(5x+1) = x+1
I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?

also why does √x-3 not equal x+9 ? isn't -3*-3 = +9?

dang is there a way to make the square root symbol extend across the numbers?

1. I assume that you want to solve

If so



Let x-3 = y. Then your equation becomes:



A product of 2 factors equals zero if one factor equals zero:



That means: x-3 = 0 ==> x = 3 or x-3=2 ==> x = 5

You must plug in the result into the original equation to check if the result is a valid solution:

x = 3:

x = 5:

Therefore: Only x = 3 is a solution.

Originally Posted by harnar

...

√(5x+1) = x+1
I can get it down to x(x-3) but then I am stuck. So yeah ok 3 works, but how was I supposed to know 0 works to?
...

1. Square both sides, you'll get:



2. A product equals zero if one factor equals zero:



Plug in the results into the original equation to check if they are valid solutions: