Solve : (sqrt(6x+7) +1)=2x?------Answer: 3----i need the steps, since i cant figure out how to do it...?

1. Add -1 to both sides:

sqrt(6x+7)=2x-1

2. Square both sides:

6x+7 = (2x-1)^2

3. FOIL out the right side:

6x+7 = 4x^2-2x-2x+1

4. Combine like terms on right:

6x+7 = 4x^2-4x+1

5. Subtract 6x and 7 from both sides:

0 = 4x^2-10x-6

6. Divide all terms by 2:

0 = 2x^2-5x-3

7. Factor polynomial:

(2x+1)(x-3)=0

8. Set each term to 0:

either 2x+1=0 or x-3=0

9. x = -1/2 or x = 3 (btw, -1/2 is not a full solution - it only satisfies the negative square root)

(i) x must satisfy the following conditions simultaneously:

6x+7≥0 → 6x≥-7 → x≥-7/6

Minimum of √(6x+7) is zero so 2x≥1+0 → x≥1/2

Both conditions become one: x≥1/2

(ii) we will rearrange and solve the equation:

√(6x+7)+1=2x → √(6x+7)=2x-1 → (√(6x+7))²=(2x-1)² → 6x+7=4x²-4x+1

→ 4x²-10x-6=0

x=(5±√(5²-4*(-6)))/4=(5±√(25+24))/4

→ x=(5±√49)/4=(5±7)/4=3 and -1/2

(iii) -1/2 is in contradiction to x≥1/2 so we eliminate -1/2 and only x=3 remains.

(iv) we should check the answer because when we square both sides of equation, some unwanted answers may appear

√(6*3+7)+1=2*3 → √(18+7)+1=6 → √25+1=6 → 5+1=6 → 6=6

(v) so x=3

sqrt(6x+7)+1=2x

Move the 1 to the right hand side of the equation...

sqrt(6x+7)=2x-1

Square both sides...

(sqrt(6x+7))^2=(2x-1)^2

Canceling the sqrt and expanding the right hand side...

6x+7=4x^2-4x+1

Subtracting the left side from the right side...

0=4x^2-10x-6

Plugging into the quadratic formula...(a=4, b=-10, c=-6)

x=(-b ± SQRT(b^2 - 4ac))/2a

Simplyfying...

x=3 or -.5...but -.5 does not satisfy the original equation, so 3 is the only solution.

√(6x+7) +1 =2x

√(6x+7) =2x-1

6x+7 = (2x-1)²

6x+7=4x²-4x+1

4x²-10x-6=0

2x²-5x-3=0

(2x+1)(x-3)=0

x=-1/2 , x=3

x=-1/2 isn't a valid solution as:

√4 + 1 ≠ -1

2≠-2

x=3 is indeed a real solution as

6=6