either x=-1/4 or x=2
and you might need to check if the MINUS form will work.
(Solution was corrected.) Answer by addingup(3677) (Show Source): You can put this solution on YOUR website! 1+sqrt(3x+3) = sqrt(7x+2)
Raise both sides to the power of 2:
(1+sqrt(3 x+3))^2 = 7x+2
Subtract 7x+2 from both sides:
-2-7x+(1+sqrt(3x+3))^2 = 0
-2-7x+(1+sqrt(3x+3))^2
= 2-4x+2sqrt(3x+3) = 0
Simplify and substitute y = sqrt(3 x+3):
2-4x+2sqrt(3x+3) = 6+2 sqrt(3x+3)-(4 sqrt(3x+3)^2)/(3)
= -(4 y^2)/3+2 y+6 = 0:
-(4 y^2)/3+2 y+6 = 0
The left hand side factors into a product with four terms:
-2/3 (y-3) (2 y+3) = 0
Multiply both sides by -3/2:
(y-3) (2 y+3) = 0
Split into two equations:
y-3 = 0 or 2 y+3 = 0
Add 3 to both sides:
y = 3 or 2 y+3 = 0
Substitute back for y = sqrt(3 x+3):
sqrt(3 x+3) = 3 or 2 y+3 = 0
Raise both sides to the power of two:
3 x+3 = 9 or 2 y+3 = 0
Subtract 3 from both sides:
3 x = 6 or 2 y+3 = 0
Divide both sides by 3:
x = 2 or 2 y+3 = 0
Subtract 3 from both sides:
x = 2 or 2 y = -3
Divide both sides by 2:
x = 2 or y = -3/2
Substitute back for y = sqrt(3 x+3):
x = 2 or sqrt(3 x+3) = -3/2
Raise both sides to the power of two:
x = 2 or 3 x+3 = 9/4
Subtract 3 from both sides:
x = 2 or 3 x = -3/4
Divide both sides by 3:
x = 2 or x = -1/4
1+sqrt(3 x+3) => 1+sqrt(3+(3 (-1))/4) = 5/2
sqrt(7 x+2) => sqrt(2+(7 (-1))/4) = 1/2:
So this solution is incorrect
1+sqrt(3 x+3) => 1+sqrt(3+3 2) = 4
sqrt(7 x+2) => sqrt(2+7 2) = 4:
So this solution is correct
The solution is:
Answer: |
| x = 2 Answer by MathTherapy(10556) (Show Source): You can put this solution on YOUR website!
What is the solution of this equation?
1+sqrt(3x+3)=sqrt(7x+2)
------ Squaring both sides ------ Squaring both sides
0 = (x - 2)(4x + 1) OR (ignore as this is an EXTRANEOUS SOLUTION