Solve. sqrtx-1=x-3?

Question

sqrtx-1=x-3

I need help

Answer 1

sqrtx-1=x-3
sqrt x = x-2
Square both sides getting:
x = x^2-4x+4
x^2-5x +4 = 0
(x-4)(x-1)=0
x= 4 or x = 1

Answer 2

First off add 1 to both sides, to get the sqrt by itself.
sqrt(x)-1+1=x-3+1
sqrt(x) = x-2. Now square both sides.
(sqrt(x))^2=(x-2)^2
x=(x-2)(x-2). Now expand using "foil"
x=x^2-4x+4. Subtract x from both sides.
x-x=x^2-4x-x+4. Combine like terms.
0=x^2-5x+4. Factor the trinomial.
0=(x-4)(x-1). Solve each term for 0
x-4=0 or x-1=0
x=4 or x=1

Answer 3

Ans =1. sqrtx+(x-3)/2 sqrt x=0 or, 2x+(x-3)=0 (x not equals to 0) or,x=1

Answer 4

sqrt (x - 1) = x - 3; is that written properly?

square both sides of the equation

x - 1 = x^2 - 6x + 9
0 = x^2 - 7x + 10
0 = (x - 5)(x - 2)
x = 2 and 5

Answer 5

x-1 = (x-3)sq
x-1 = xsq - 6x + 9
0 = xsq -7x + 10
0 = (x-5)(x-2)
x=5 x=2