please help 3+√(3x+1)=x?
ive been trying to solve it i isolated the the radical then raise both sides to the power of 2 then i get confuse the book says the answer is 8 can some explain this if possible p.s everything in parenthesis is under the radical
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3 + √(3x + 1) = x
√(3x + 1) = x - 3
3x + 1 = (x - 3)²
3x + 1 = x² - 6x + 9
x² - 6x - 3x + 9 - 1 = 0
x² - 9x + 8 = 0
(x - 8)(x - 1) = 0
If the product of two factors equals zero, then one or both factors equal zero.
If x - 8 = 0,
x = 8
If x - 1 = 0,
x = 1
x { 1, 8 }
¯¯¯¯¯¯¯¯
x - 3 = â(3x + 1)
(x - 3)^2 = 3x + 1
x^2 - 6x + 9 = 3x + 1
x^2 - 9x + 8 = 0
(x-8)(x-1) = 0
x = 1, 8
These are supposedly the solutions, but we must check them because squaring the side potentially creates solutions that aren't really solutions.
3 + â(4) = 1
3 + 2 = 1 [That doesn't work, so x = 1 is not a solution]
3 + â(24+1) = 8
3 + â(25) = 8
3 + 5 = 8 [That works, so x = 8 is a solution]
So, the solution is x = 8. There you are.
rearrange which is what i think you were doing:
3+â(3x+1) = x
â(3x+1) = x-3
3x+1 = (x-3)^2
3x+1 = x^2 -6x +9
x^2 -9x +8 = 0
then factorise it using the quadratic formula and one of the answers should be 8:
9+-â((-9)^2 -(4*1*8)) / 2*1
9+-â(81-32) / 2
9+-â(49) /2
(9+7) /2 = 8
(9-7) /2 =1
V(3x+1)=x-3
[V(3x-1)]²=(x-3)²
3x+1=x²-6x+9
x²-6x-3x+9-1=0
x²-9x+8=0
Aplying Bhaskara
x'=8
x"=1
I am from Brazil