SOLUTION: Solve each equation and to check for extraneous solutions [√(2x^2 + 5x + 6)]=x

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Question 1185300: Solve each equation and to check for extraneous solutions
[√(2x^2 + 5x + 6)]=x

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

sqrt%282x%5E2+%2B+5x+%2B+6%29%22%22=%22%22x

We square both sides:

%28sqrt%282x%5E2+%2B+5x+%2B+6%29%29%5E2%22%22=%22%22%28x%29%5E2

When we square a square root we take away the square and the square root:

2x%5E2+%2B+5x+%2B+6%22%22=%22%22x%5E2

Subtract x2 from both sides:

x%5E2+%2B+5x+%2B+6%22%22=%22%220

%28x%2B3%29%28x%2B2%29%22%22=%22%220

Use the zero factor principle. Set each factor equal to 0:

x+3 = 0;   x+2 = 0
  x = -3;    x = -2

We substitute those in the original equation to see if both are
solutions to the original equations, or extraneous:

Substituting x = -3

sqrt%282%28-3%29%5E2+%2B+5%28-3%29+%2B+6%29%22%22=%22%22-3

sqrt%282%289%29+-+15+%2B+6%29%22%22=%22%22-3

sqrt%2818+-+15+%2B+6%29%22%22=%22%22-3

sqrt%289%29%22%22=%22%22-3

3%22%22=%22%22-3

That's false.  So -3 is an extraneous solution, so we discard it.

Substituting -2

sqrt%282%28-2%29%5E2+%2B+5%28-2%29+%2B+6%29%22%22=%22%22-2

sqrt%282%284%29+-+10+%2B+6%29%22%22=%22%22-2

sqrt%288+-+10+%2B+6%29%22%22=%22%22-2

sqrt%284%29%22%22=%22%22-2

2%22%22=%22%22-2

That's false.  So -2 is also an extraneous solution, so we discard it.

The equation has no solution.

Note: The symbol "√(" always means the POSITIVE square root. To indicate
the negative square root, there must be a negative sign before it "-√(".

Edwin