1. ## Another confusing radicals problem

root(2x +5) + 2root(x+6) = 5

2. Originally Posted by Trentt
root(2x +5) + 2root(x+6) = 5
Square:

(2x+5) + 4 sqrt(2x+5) sqrt(x+6) +4 (x+6) = 25

so:

4 sqrt(2x+5) sqrt(x+6) = -4-6x.

Now square again:

16 (2x+5)(x+6) = (4+6x)^2

or:

32·x^2 + 272·x + 480 = 36·x^2 + 48·x + 16

Which simplifies to:

4·(x^2 - 56·x - 116)=0

This is a quadratic with roots x=58, and x=-2.

Now we have to substitute each of these back into the original equation
to check that they are roots (as the squarings can introduce spurious
roots).

root(2 58 +5) + 2root(58+6) = 27,

so 58 is not a root of the original equation.

root(2(-2) +5) + 2root((-2)+6) = 1+2 (2)=5

so x=-2 is a root.

Therefore the only solution to the original equation is x=-2.

RonL

3. Originally Posted by CaptainBlack
Which simplifies to:

4·(x^2 - 56·x - 116)=0

This is a quadratic with roots x=58, and x=-2.
So what happens to the bold underlined 4?

4. Hello, Trentt!

root(2x +5) + 2·root(x+6) .= .5

We have: .2·root(x + 6) .= .5 - root(2x + 5)

Square both sides: .[2·root(x + 6)]˛ .= .[5 - root(2x + 5)]˛

. . 4(x + 6) .= .10 - 10·root(2x + 5) + 2x + 5

. . 2x + 14 .= .-10·root(2x + 5)

Divide by 2: .x + 7.= .-5·root(2x + 5)

Square both sides: .(x + 7)˛ .= .[-5·root(2x + 5)]˛

. . x˛ + 14x + 49 .= .25(2x + 5)

which simplifies to: .x˛ - 36x - 76 .= .0

which factors: .(x + 2)(x - 38) .= .0

and has roots: .x = -2, 38

But only x = -2 satisfies the equation.