solve. The square root of x -1 = x - 3?

Question

solve.
1) the square root of x -1 = x - 3

2)the square root of y +12 = 2

Answer 1

1) sqrt(x - 1) = x - 3
Square both sides:
x - 1 = (x - 3)²
x - 1 = x² -6x + 9
x² -7x + 10 = 0
(x - 2)(x - 5) = 0
x = 2 or x = 5

As a double-check:
sqrt(2 - 1) = 2 - 3
sqrt(1) = -1 <-- this is a true statement since 1 and -1 are roots.

sqrt(5 - 1) = 5 - 3
sqrt(4) = 2 <-- check

2) sqrt(y + 12) = 2
Square both sides:
y + 12 = 2²
y + 12 = 4
y = -8

As a double-check:
sqrt(-8 + 12) = 2
sqrt(4) = 2 <-- correct

So your answers are:
1) x = 2 or x = 5
2) y = -8

Answer 2

I'll do the first one.

Take the square of both sides so you have x - 1 = (x-3)^2
Multiply out the (x-3)^2 = x^2 - 6x + 9

Therefore, x^2 - 6x + 9 = x - 1, set up to equal 0 for a quadratic equation:

Subtract x from both sides: x^2 - 7x + 9 = -1

Add 1 to both sides: x^2 - 7x + 10 = 0

Factor: (x-2)(x-5)

X = 2 or 5

However, we may have introduced another number when we squared both sides so CHECK:

For 2: squareroot (2 - 1) = 1 , but 2 - 3 = -1. Not an answer,
For 5: squareroot (5 - 1) = 2, 5 - 2 also 2.

Answer is 5.

Answer 3

It is not clear from your question if you are saying sq. rt. of (x)-1,or sq. rt. of (x-1) I am assuming it is the latter.
As always let's try to get x to stand alone. The way I start out with this type of problem would be to get rid of the sq. rt. sign. We do this by squaring both sides of the equation.
sq. rt. (x-1)= x-3, after squaring we get x-1 = (x-3)(x-3),after multiplying the terms we get,
x-1= x squared -6x+9 Now we want to get zero on one side of the equation and everything else on the other side.
0= x squared-7x+10. Now we can factor to find out what x could be. (x-2)(x-5) works. Therefore x = 2 or, x=5

The next problem works the same way, but it is easier. We Still need to square both sides in order to end up having our y stand alone.
The sq.rt. of (y+12) = 2, after squaring you have y+12 =4. To end up with y all by itself we must subtract 12 from both sides of the equation. (y+12) -12 = 4 -12, y = -8
I try to plug the number I get into the original equation to make sure it's correct. Hope this helps.

Answer 4

1)
A) square root of x-1=x-3
B) square both sides to get x-1=(x-3)^2
C) do the math x-1=x^2-6x+9
D) Make the left side =0 0=x^2-7x+10
E) foil (x-2)(x-5)=0
F) so the solutions are x=2 and 5

2)
A) square root y+12=2
B) square both sides y+12=4
C) subtract 12 from both sides to get y by itself y=-8
D) so your solution is y=-8

Answer 5

1) sqrt(x-1) = x-3, square both sides
x-1 = (x-3)^2 = x^2 - 6x + 9

x^2 - 7x + 10 = 0, solve the quadratic equation.

Answer 6

2)y+12=2
so first subtract 12 from 2 getting -10 then take the square root of that


not sure on #1 =]

Answer 7

1. sq.rt.(x-1) = (x-3)
=> (x-1) = sq. (x-3)
=> x-1 = sq (x) -6x + 9
=> sq (x) - 7x + 10 = 0
=> (x-5) (x-2) = 0
Therefore, x = 5 and x = 2

2. sq.rt.(y+12) = 2
=> y+12 = 4
=> y-6 = 0
Therefore, y = 6

Answer 8

Given: sqrt(x-1) = x-3
square both sides: x-1 = (x-3)(x-3)
x-1 = x^2-6x+9
subtract (x-1): 0 = x^2-7x+10
factor: 0 = (x-5)(x-2)
solve: x=5 or x=2

Plug and chug to check your answer:
sqrt(5-1) = 5-3? sqrt(4) = 2 yep!
sqrt(2-1) = 2-1? sqrt(1) = 1 yep!

Answer 9

All you have to do is to square both sides..
This becomes
1.
x-1=x^2-6x+9
0=x^2-7x+10
(x-5)(x-2)=0
2 doesn't work so
x={5}

2.
y+12=4
y={-8}

Answer 10

When you say "the sqrt of x-1" are you meaning (x^1/2) - 1, or (x-1)^1/2? It will make a difference in your answers!