Check for extraneous solutions

Question

(√2x+5)=x+3

Answer 1

if I read this right, and the square root is of the quantity (2x+5)

 

SQRT(2X+5) = X + 3     square both sides

 

2X + 5 = (X+3)^2

 

2X + 5 = X² + 6X + 9     combine terms, and put all on left side

 

X² + 4X + 4 = 0

 

(X+2)(X+2) = 0            so there is a double root at X = -2

 

Check in orig eqn.

 

SQRT(2(-2) + 5)  = 1

SQRT(1) = 1                  checks

 

satisfies the eqn, and no extraneous roots, or values that would cause the sqrt argument to be negative

Answer 2

Hi Amelia,

 

Assuming this is:

 

sqrt(2x) + 5 = x + 3

 

Isolate the "sqrt(2x)" by subtracting 5 from each side.

 

sqrt(2x) = x - 2

 

Square both sides to extricate the "2x" from under the radical.

 

2x = (x-2)²

 

2x = x² -4x + 4

 

Put everything on the right, so the left =0

 

0 = x² -6x + 4

 

This solves to x = 3 ± √5

 

 

Trying (3 + √5) back into the original equation,

we get 8.23607  =  8.23607 which is true.

So the solution (3 + √5) works.

 

 

Trying (3 - √5) back into the original equation,  

we get  6.23607 = 3.763932 which is NOT true.

 

So (3 - √5) does NOT work, so it is the extraneous solution.