SOLUTION: Please help Solve: (sqrt of 2x - 3) = x - 1 Solve: (sqrt of 3x - 3) - (sqrt of x) = 1

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Question 728593: Please help
Solve: (sqrt of 2x - 3) = x - 1

Solve: (sqrt of 3x - 3) - (sqrt of x) = 1

Found 3 solutions by MathLover1, MathTherapy, greenestamps:
Answer by MathLover1(20855) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt%282x+-+3%29+=+x+-+1....square both sides
%28sqrt%282x+-+3%29%29%5E2+=+%28x+-+1%29%5E2
2x+-+3+=+x%5E2-2x+%2B1
0=+x%5E2-2x-2x%2B3+%2B1
0=+x%5E2-4x%2B4...use quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%2A1%2A4+%29%29%2F%282%2A1%29+
x+=+%284+%2B-+sqrt%28+16-16+%29%29%2F2+
x+=+%284+%2B-+sqrt%28+0+%29%29%2F2+
x+=+%284+%2B-+0%29%2F2+
so, solution is
x+=+4%2F2+
x+=+2+

choice: (c.) 2


sqrt%283x-3%29+-+sqrt%28x%29+=+1
sqrt%283x-3%29++=+sqrt%28x%29%2B1...square both sides
%28sqrt%283x-3%29%29%5E2+=+%28sqrt%28x%29%2B1%29%5E2
3x-3+=x%2B2sqrt%28x%29%2B1
3x-x-3-1+=2sqrt%28x%29
2x-4+=2sqrt%28x%29
2%28x-2%29+=2sqrt%28x%29
cross%282%29%28x-2%29+=cross%282%29sqrt%28x%29
x-2+=sqrt%28x%29..square both sides again
%28x-2%29%5E2+=%28sqrt%28x%29%29%5E2
x%5E2-4x%2B4+=x

x%5E2-4x-x%2B4+=0
x%5E2-5x%2B4+=0

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%2A1%2A4+%29%29%2F%282%2A1%29+

x+=+%285+%2B-+sqrt%2825-16+%29%29%2F2+
x+=+%285+%2B-+sqrt%289+%29%29%2F2+

x+=+%285+%2B-+3%29%2F2+
solutions:
x+=+%285+%2B3%29%2F2+
x+=+8%2F2+
x+=+4+
and
x+=+%285+-3%29%2F2+
x+=+2%2F2+
x+=+1+

choice: (c.) 1 and 4

Answer by MathTherapy(10858) About Me  (Show Source):
You can put this solution on YOUR website!
Please help

Solve: (sqrt of 2x - 3) = x - 1

Solve: (sqrt of 3x - 3) - (sqrt of x) = 1
**********************************
Again, respondent @MathLover1(20855) used the QUADRATIC EQUATION formula to solve both quadratic equations,
which is UNNECESSARY, since the resultant TRINOMIALS are FACTORABLE! 

Furthermore, for sqrt%283x+-+3%29+-+sqrt%28x%29+=+1, she got 2 values for x: x = 4, and x = 1. Obviously, when checked, the value 1 DOES
NOT SATISFY this equation, thereby making 1, an EXTRANEOUS solution. So, in this case, x = 4 is the SOLE VALID and
ACCEPTABLE SOLUTION.

Answer by greenestamps(13367) About Me  (Show Source):
You can put this solution on YOUR website!


You can get good mental exercise by solving these equations using logical reasoning and simple mental arithmetic.

Start by assuming (hoping) that the solutions are integers. Then use logical reasoning to determine the solution(s).

(1) sqrt%282x-3%29=x-1

In this equation, we want (2x-3) to be a perfect square. And if x is an integer, the (2x-3) is an odd perfect square. So look at values of x for which (2x-3) is an odd perfect square.

(a) 2x-3 = 1; 2x = 4; x = 2. sqrt(2x-3) = sqrt(1) = 1; and 2-1 = 1. So x = 2 is a solution.

(b) 2x-3 = 9; 2x = 12; x = 6. sqrt(2x-3) = 3 but x-1 = 6. x = 6 is not a solution.

For larger values of x, "x-1" will increase faster than "sqrt(2x-3)", so there are no more solutions.

ANSWER: x = 2

(2) sqrt%283x-3%29-sqrt%28x%29=1

In this equation, we want x and (3x-3) to both be perfect squares.

(a) x = 1; 3x-3 = 0; both are perfect squares. However, trying x = 1 in the equation gives us "0-1 = 1", which is not true. Note that the two sides of the equation evaluate to numbers that are opposites. This is a common occurrence in equations like this, where the formal solution process including squaring both sides of an equation leads to extraneous solutions.

So x = 1 is NOT a solution.

(b) x = 4; 3x-3 = 9 is also a perfect square; and sqrt(3x-3) = 3 and x-1 = 3, so x = 4 is a solution.

Similarly to the first problem, for larger values of x "sqrt(3x-3)" will increase faster than "sqrt(x)", so again there are no more solutions.

ANSWER: x = 4