How do you solve $x = sqrt(-3x + 10)$ and find any extraneous solutions?

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Answer 1 · 2016-07-09T09:37:09

$x=-5" and "x=+2" "$ are solutions to $x^2+3x-10=0$

Extraneous $ul("values")$ of $x>10/3$

For $AAx notin {-5,2}" " =>" " x^2+3x-10 != 0$

$color(blue)("Determine general solution")$

Square both sides

$x^2=-3x+10" "larr" you now have a quadratic"$

$x^2+3x-10=0$

Notice that $2xx5=10" and "5-2=+3$

Thus the factorization is $" "(x+5)(x-2)=0$

So $x=-5" and "x=+2$ are general solutions
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$color(blue)("Determine extraneous values of "x)$

For the number to remain in the $ul("'Real'")$ domain you may can not permit the square root to be that of a negative number.

So $-3x+10>=0$

Multiply by (-1) giving

$+3x-10<=0$
Notice that the inequality sign has turned the other way. This happens when you multiply by (-1)

Add 10 to both sides

$3x<=10$

$=>x<=10/3$ as a required condition

$=>x>10/3$ takes the solution out of the Real domain and into the Complex number domain

How do you solve x = sqrt(-3x + 10) x = sqrt(-3x + 10) and find any extraneous solutions?