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How do you solve and check for extraneous solutions in $sqrt(x+1) + 5 = x$ ?

1 Answer

Aug 1, 2015

The only valid solution is $x=8$
A second candidate solution $x=3$ can be eliminated by checking for the validity of the given equation with $x=3$ (and noting that it is not valid).

Explanation:

Given $sqrt(x+1) + 5 = x$

Subtract $5$ from both sides
$color(white)("XXXX")$ $sqrt(x+1) = x-5$

Square both sides (possibly generating an extraneous root at this point)
$color(white)("XXXX")$ $x+1 = x^2-10x+25$

Subtract $(x+1)$ from both sides (and flip sides)
$color(white)("XXXX")$ $x^2-11x+24 = 0$

Factor
$color(white)("XXXX")$ $(x-3)(x-8) = 0$

$rArr$ $color(white)("XXXX")$ $x=3$ $color(white)("XXXX")$ or $color(white)("XXXX")$ $x=8$

Substituting $3$ for $x$ in the Left Side of original equation
$color(white)("XXXX")$ $sqrt(3+1)+5 = 2+3$
$color(white)("XXXX")$ $color(white)("XXXX")$ $!=3$
The solution $x=3$ is extraneous.

Substituting $8$ for $x$ in the Left Side of original equation
$color(white)("XXXX")$ $sqrt(8+1)+5 = 3+5$
$color(white)("XXXX")$ $color(white)("XXXX")$ $=8$
$color(white)("XXXX")$ $color(white)("XXXX")$ $=x$
The solution $x=8$ is valid.

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