How do you solve the equation: #x=sqrt(3x+40)#?

Question

Ethan Adkins · Accepted Answer

#x=8#

#x = sqrt{3x + 40}#

Over reals the radical sign refers to the principal square root, so squaring both sides may introduce extraneous solutions.

#x^2 = 3x+40#

#x^2 - 3x -40 = 0#

#(x-8)(x+5)=0#

#x=8 or x=-5#

The principal square root won't ever be #-5# so we reject that as extraneous.

#x=8#

Check:

# sqrt{3(8)+40}=sqrt{64}=8 quad sqrt#

Avery Alexander · Answer

undefined To solve the equation x = √(3x + 40), we can follow these steps:

1. Start by isolating the square root term on one side of the equation. Square both sides of the equation to eliminate the square root: (x)^2 = (√(3x + 40))^2.

2. Simplify the equation: x^2 = 3x + 40.

3. Rearrange the equation to bring all terms to one side: x^2 - 3x - 40 = 0.

4. Factorize the quadratic equation: (x - 8)(x + 5) = 0.

5. Set each factor equal to zero and solve for x: x - 8 = 0 or x + 5 = 0.

6. Solve for x in each equation: x = 8 or x = -5.

Therefore, the solutions to the equation x = √(3x + 40) are x = 8 and x = -5.