Thread: couple math 95 problems

1. 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?

2. 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

$x-2\sqrt{x-3}=3$ If so

$-2\sqrt{x-3} =3-x~\implies~2\sqrt{x-3}=x-3$

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

$-2y = y^2~\implies~0=y^2-2y~\implies~y(y-2)=0$

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

$y = 0~\vee~y-2=0~\implies~y=0~\vee~y=2$

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: $3-2\sqrt{3-3}=3$

x = 5: $5 - 2\sqrt{5-3} \neq 3$

Therefore: Only x = 3 is a solution.

3. 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:

$x^2-3x=0~\implies~x(x-3)=0$

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

$x = 0~\vee~x-3=0~\implies~\boxed{x=0~\vee~x=3}$

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

$x = 0: \sqrt{0+1} = 0+1~\longrightarrow~OK$

$x = 3: \sqrt{15+1} = 3+1~\longrightarrow~OK$