+0

# sqrt(2x+3)+2= sqrt(6x+7) I have two radicals that equal each other and i need to find x, the only thing your calculator does is take my typed in variables and

0
95
1

Guest Mar 5, 2017
Sort:

#1
0

Solve for x:
2 + sqrt(2 x + 3) = sqrt(6 x + 7)

Raise both sides to the power of 2:
(2 + sqrt(2 x + 3))^2 = 6 x + 7

Subtract 6 x + 7 from both sides:
-7 - 6 x + (2 + sqrt(2 x + 3))^2 = 0

-7 - 6 x + (2 + sqrt(2 x + 3))^2 = 4 sqrt(2 x + 3) - 4 x:
4 sqrt(2 x + 3) - 4 x = 0

Simplify and substitute y = sqrt(2 x + 3):
4 sqrt(2 x + 3) - 4 x = 6 + 4 sqrt(2 x + 3) - 2 (sqrt(2 x + 3))^2 = -2 y^2 + 4 y + 6 = 0:
-2 y^2 + 4 y + 6 = 0

The left hand side factors into a product with three terms:
-2 (y - 3) (y + 1) = 0

Divide both sides by -2:
(y - 3) (y + 1) = 0

Split into two equations:
y - 3 = 0 or y + 1 = 0

y = 3 or y + 1 = 0

Substitute back for y = sqrt(2 x + 3):
sqrt(2 x + 3) = 3 or y + 1 = 0

Raise both sides to the power of two:
2 x + 3 = 9 or y + 1 = 0

Subtract 3 from both sides:
2 x = 6 or y + 1 = 0

Divide both sides by 2:
x = 3 or y + 1 = 0

Subtract 1 from both sides:
x = 3 or y = -1

Substitute back for y = sqrt(2 x + 3):
x = 3 or sqrt(2 x + 3) = -1

Raise both sides to the power of two:
x = 3 or 2 x + 3 = 1

Subtract 3 from both sides:
x = 3 or 2 x = -2

Divide both sides by 2:
x = 3 or x = -1

2 + sqrt(2 x + 3) ⇒ 2 + sqrt(3 + 2 (-1)) = 3
sqrt(6 x + 7) ⇒ sqrt(7 + 6 (-1)) = 1:
So this solution is incorrect

2 + sqrt(2 x + 3) ⇒ 2 + sqrt(3 + 2 3) = 5
sqrt(6 x + 7) ⇒ sqrt(7 + 6 3) = 5:
So this solution is correct

The solution is: