Question

Solve the radical equation and what is the extraneous solution to the radical equation

Answer 1

Answer:

x = -6 and x= -1

Step-by-step explanation:

the expression we have is:

$x+4=\sqrt{x+10}$

Taking tha square of the whole equation:

$(x+4)^2=x+10$

solving the binomial squared on the left side with the rule:

$(a+b)^2=a^2+2ab+b^2$

we get:

$x^2+8x+16=x+10$

rearranging all terms to the left:

$x^2+8x-x+16-10=0$

joining like terms:

$x^2+7x+6=0$

we can sove this quadratic equation with the quadratic formula, or by factoring:

$x^2+7x+6=(x+6)(x+1)=0$

and from this we find our two solutions using the zero product property (if a product of thing is equal to zero, one of them must be zero)

$x+6=0\\x=-6$

and

$x+1=0\\x=-1$

Answer 2

Remove the radical by raising each side to the index of the radical.

x=-1