Question

I know that I posted this already but my question has not been answered..

Solve in the exact form.

Original problem:
(sqrt of 4x+1)+(sqrt of x+1)=2

4x+1 + 2√(4x^2+13x+3) + x+3 = 4
Where does the 2 come from according to the equation above?? This 2:-----> 2√(4x^2+13x+3)

Answer 1

Hi!
the question is
¡Ô(4x+1) + ¡Ô(x+1) = 2

Square both sides
(¡Ô(4x+1) + ¡Ô(x+1))^2 = 2^2

still remember the identity
(a+b)^2 = a^2 + 2ab + b^2
let ¡Ô(4x+1) be a
¡Ô(x+1) be b
as a^2 + 2ab + b^2
we can derived the equation
4x+1 + 2¡Ô(4x^2+13x+3) + x+3 = 4

Answer 2

√(4x+1) = 2 - √(x+1)
square both sides
(√(4x+1) )^2 = (2 - √(x+1) )^2
4x+1 = 4 - 4√(x+1) + x+1
4√(x+1) = 4 - 3x
square both sides again
16(x+1) = 16 - 24x + 9x^2
16x + 16 = 9x^2 - 24x + 16
9x^2 - 40x = 0
x(9x-40) = 0
x = 0 or x = 40/9

HOWEVER, since we squared the equation, all answers must be verfied in the original equation
√(4x+1) + √(x+1) = 2

if x = 0
LS = √(0+1) + √(0+1)
= 1+1 = 2
= RS

if x = 40/9
LS =√(160/9 + 1) + √(40/9 + 1)
= √(169/9) + √(49/9)
= 13/3 + 7/3
= 20/3
≠ RS

So the only solution is x = 0

check with Wolfram :
http://www.wolframalpha.com/input/?i=solve+√%284x%2B1%29+%3D+2+-+√%28x%2B1%29