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
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.