Question

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b) \\(sqrt{2x + 3} = x\\)
c) \\(x = sqrt{x + 4} + 2\\)

Explanation:

Part (b)

Step 1: Square both sides to eliminate the square root

To solve the equation \(\sqrt{2x + 3}=x\), we square both sides. Squaring the left side \((\sqrt{2x + 3})^2 = 2x+3\) and squaring the right side \(x^2\). So we get the equation:
\(2x + 3=x^2\)

Step 2: Rearrange into standard quadratic form

Rearrange the equation \(2x + 3=x^2\) to the standard quadratic form \(ax^2+bx + c = 0\). Subtract \(2x\) and \(3\) from both sides:
\(x^2-2x - 3=0\)

Step 3: Factor the quadratic equation

Factor the quadratic \(x^2-2x - 3\). We need two numbers that multiply to \(- 3\) and add to \(-2\). The numbers are \(-3\) and \(1\). So,
\(x^2-2x - 3=(x - 3)(x+1)=0\)

Step 4: Solve for \(x\)

Set each factor equal to zero:

• \(x - 3=0\) gives \(x = 3\)
• \(x + 1=0\) gives \(x=-1\)

Step 5: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \(x = 3\): Left side \(\sqrt{2(3)+3}=\sqrt{6 + 3}=\sqrt{9}=3\), Right side \(=3\). So \(x = 3\) is a valid solution.
• For \(x=-1\): Left side \(\sqrt{2(-1)+3}=\sqrt{-2 + 3}=\sqrt{1}=1\), Right side \(=-1\). So \(x=-1\) is an extraneous solution.

Part (c)

Step 1: Isolate the square root

Given the equation \(x=\sqrt{x + 4}+2\), we isolate the square root by subtracting \(2\) from both sides:
\(x-2=\sqrt{x + 4}\)

Step 2: Square both sides to eliminate the square root

Square both sides. The left side \((x - 2)^2=x^2-4x + 4\) and the right side \((\sqrt{x + 4})^2=x + 4\). So we get:
\(x^2-4x + 4=x + 4\)

Step 3: Rearrange into standard quadratic form

Subtract \(x\) and \(4\) from both sides:
\(x^2-5x=0\)

Step 4: Factor the quadratic equation

Factor out an \(x\) from \(x^2-5x\):
\(x(x - 5)=0\)

Step 5: Solve for \(x\)

Set each factor equal to zero:

• \(x=0\)
• \(x - 5=0\) gives \(x = 5\)

Step 6: Check for extraneous solutions

Check these solutions in the original equation.

• For \(x = 0\): Left side \(=0\), Right side \(\sqrt{0 + 4}+2=\sqrt{4}+2=2 + 2=4\). So \(x = 0\) is an extraneous solution.
• For \(x = 5\): Left side \(=5\), Right side \(\sqrt{5 + 4}+2=\sqrt{9}+2=3 + 2=5\). So \(x = 5\) is a valid solution.

Answer:

s:

• For part (b): \(x = 3\)
• For part (c): \(x = 5\)