Solve √(x-1)=x-3?

Question

I got x=x^2-6x+10 but not sure if Im supposed to go any further!

Answer 1

Square both sides:
x-1=(x-3)^2
x-1=x^2-6x+9
x^2-7x+10=0
(x-2)(x-5)=0
x=2 and 5
5 would be the answer because both sides will equal 2. 2 would not be included in the answer because the left hand side would equal 1 and the right hand side would equal -1.

Answer 2

i am going to do the first one. Take the sq. of both area so that you've x - a million = (x-3)^2 Multiply out the (x-3)^2 = x^2 - 6x + 9 therefore, x^2 - 6x + 9 = x - a million, set up to equivalent 0 for a quadratic equation: Subtract x from both area: x^2 - 7x + 9 = -a million upload a million to both area: x^2 - 7x + 10 = 0 ingredient: (x-2)(x-5) X = 2 or 5 in spite of the undeniable fact that, we may have presented yet another huge form when we squared both area so examine: for 2: squareroot (2 - a million) = a million , yet 2 - 3 = -a million. no longer an answer, for 5: squareroot (5 - a million) = 2, 5 - 2 also 2. answer is 5.

Answer 3

raise to the second power

x-1 = x^2 - 6x + 9

x^2 -7x + 10 = 0

(x - 2)(x - 5) = 0

There are two solutions to the quadrant equation:

x=2, but it is not a solution for the original equation because
x-3 = -1, and sqrt(2 - 1)=sqrt(1)=1 (The value returned from the sqrt function is defined to be positive)

x=5, let's put it in the equation:

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

This is the solution.

Answer 4

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

check the answer and you see that 2 can't be the answer.

therefore x=5

Answer 5

x-1 = x^2 - 6x + 9
-10 = x^2 - 7x
x^2 - 7x + 10 = 0
-------------------------------------------------
Quadratic Equation:
x=(-b ± √(b^2 - 4ac)) / 2a
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x = (-(-7) ± √[(-7^2) - 4(1)(10)]) / 2(1)
x = (7 ± √[(49) - 40]) / 2
x = (7 ± √9) / 2
x = (7 + 3) / 2 or x = (7 - 3) / 2
x = 10/2 or x = 4/2
x = 5 or x = 2

If you plug 5 or 2 in for "x" in the original equation, you'll see that only 5 works:

√(x-1) = x - 3
√(5-1) = 5 - 3
√4 = 2
2 = 2
"5" works!

√(x-1) = x - 3
√(2-1) = 2 - 3
√1 = -1
1 ≠ -1
"2" does not work.

Answer 6

first, square both sides to remove sqrt sign.

x-1 = (x-3)^2
x-1 = x^2 - 6x + 9
x^2 - 6x + 9 = x-1
x^2 - 7x + 10 = 0

solve for x

factor
(x-5)(x-2) = 0
x = 5 or x = 2

Answer 7

goin right
solve it and dnt forget to verify your answer as you squared the equation.
both roots must be greater than 1 else the under root term will become imaginary!

Answer 8

sq.root(x-1)=x-3
or, x-1=(x-3)^2
or, x-1=x^2-6x+9
or, x^2-6x-x+9+1=0
or, x^2-7x+10=0
or, x^2-5x-2x+10=0
or, x(x-5)-2(x-5)=0
or, (x-5)(x-2)=0

Therefore, x= 5 or 2