Anyone explain how to solve. also ignore the work, !

9/2+sqrt(21)/2 =  x

Step-by-step explanation:

sqrt(x+1) = x-4

Square each side

(sqrt(x+1))^2  = (x-4)^2

x+1 = (x-4) (x-4)

FOIL

x+1 = x^2 -4x-4x+16

Combine like terms

x+1 = x^2 -8x+16

Subtract x from each side

x-x+1 = x^2 -8x-x+16

1 = x^2 -9x+16

Subtract 16 from each side

1-16 = x^2 -9x+16-16

-15=x^2 -9x

Completing the square

(-9/2)^2 = 81/4

Add 81/4

-15 +81/4 = x^2 -9x +81/4

Get a common denominator

-60/4 +81/4 = (x-9/2)^2

21/4 =  (x-9/2)^2

Take the square root of each side

±sqrt(21/4) = sqrt( (x-9/2)^2)

±sqrt(21/4) =  (x-9/2)

We know the sqrt(a/b) =sqrt(a)/sqrt(b)

±sqrt(21)/sqrt(4) =  (x-9/2)

±sqrt(21)/2 =  (x-9/2)

Add 9/2 to each side

9/2±sqrt(21)/2 =  x-9/2+9/2

9/2±sqrt(21)/2 =  x

We need to check the solutions since we squared as our first step and can get extraneous solutions

sqrt(9/2+sqrt(21)/2+1) = 9/2+sqrt(21)/2-4

This is a valid solution

sqrt(9/2-sqrt(21)/2+1) = 9/2-sqrt(21)/2-4

 The right side is negative  and the left side is a square root.  This cannot happen, so it is an extraneous solution

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