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Solve the equation $x = \sqrt{x + 6}$, $x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 5 - 2015

Question 5

$Solve-the-equation-$x-=-\sqrt{x-+-6}$,-$x-\in-\mathbb{R}$.-Leaving Cert Mathematics-Question 5-2015.png$

Solve the equation $x = \sqrt{x + 6}$, $x \in \mathbb{R}$.

Worked Solution & Example Answer:Solve the equation $x = \sqrt{x + 6}$, $x \in \mathbb{R}$. - Leaving Cert Mathematics - Question 5 - 2015

Step 1

Step 1: Rearranging the Equation

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Answer

Starting with the equation:\n\n $x = \sqrt{x + 6}$ \n\nSquare both sides to eliminate the square root:\n\n $x^2 = x + 6$

Step 2

Step 2: Forming a Quadratic Equation

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Answer

Rearranging gives us a standard quadratic form:\n\n $x^2 - x - 6 = 0$

Step 3

Step 3: Factoring the Quadratic

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Answer

This can be factored as:\n\n $(x - 3)(x + 2) = 0$

Step 4

Step 4: Finding the Roots

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Answer

Setting each factor to zero gives us the possible solutions:\n\n $x - 3 = 0 \Rightarrow x = 3$ \n $x + 2 = 0 \Rightarrow x = -2$

Step 5

Step 5: Checking for Extraneous Solutions

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Answer

We need to check each solution by substituting back into the original equation:\n\nFor $x = 3$ : \n $3 = \sqrt{3 + 6} = \sqrt{9} = 3$ (Valid)\n\nFor $x = -2$ : \n $-2 = \sqrt{-2 + 6} = \sqrt{4} = 2$ (Not valid)\n\nThus, the only solution is $x = 3$ .

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