Solve for x the equation x+2=squareroot(2x+12)?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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The first step is to impose constraint of existence of the radical.

2x+12 `>=` 0

We'll isolate 2x to the left side:

2x `>=` -12

x `>=` -6

The values of x that can be validated as solutions of the given equation must belong to the closed interval [-6,+`oo` ).

Now, we'll solve the equation raising to square both sides, to remove the radical:

(x+2)^2 = 2x + 12

We'll expand the binomial:

x^2 + 4x + 4 = 2x + 12

We'll shift all terms to the left side:

x^2 + 2x - 8 = 0

We'll apply quadratic formula:

x1 = [-b+sqrt(b^2 - 4ac)]/2a

a=1 , b = 2 and c = -8

x1 = [-2+sqrt(4 + 32)]/2

x1 = (-2+6)/2

x1 = 2

x2 = (-2-6)/2

x2 = -4

Since both values of x are located in the closed interval [-6 , `oo` ), the solutions of the equation are {-4 ; 2}.

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