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

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Answer 1 · 2016-09-02T13:27:52

Solution is $x=5$ ( $x=-2$ is extraneous solution).

Here we can not have $3x+10<0$ i.e. $x<-10\3$ and any value of $x<-10/3$ is extraneous.

Squaring both sides in $sqrt(3x+10)=x$ , we get

$3x+10=x^2$ or

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

$x^2-5x+2x-10=0$ or

$x(x-5)+2(x-5)$ or

$(x+2)(x-5)=0$ i.e.

either $x+2=0$ i.e. $x=-2$

or $x-5=0$ i.e. $x=5$

As none of these id less than $-10/3$ ,

Solution is $x=-2$ or $x=5$ , but $x=-2$ id extraneous as putting $x=-2$ in given equation gives us $2=-2$ .

Hence solution is $x=5$ .

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