Question

solve for x, where x is a real number.\\(\sqrt{3x + 16} - \sqrt{x + 13} = 1\\)\\(\\) (if there is more than one solution, separate them with commas.)\\(x = \\)\\(\\) no solution \\(\sqrt{\square}\\) \\(\sqrt\square{\square}\\) \\(\frac{\square}{\square}\\) \\(\square,\square,\dots\\)

Explanation:

Step1: Isolate one square root

Start with the equation \(\sqrt{3x + 16}-\sqrt{x + 13}=1\). Add \(\sqrt{x + 13}\) to both sides to isolate \(\sqrt{3x + 16}\):
\(\sqrt{3x + 16}=1+\sqrt{x + 13}\)

Step2: Square both sides

Square both sides of the equation to eliminate the square root on the left. Recall that \((a + b)^2=a^2 + 2ab + b^2\), so:

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Step3: Simplify the equation

Simplify the right - hand side: \(1+(x + 13)=x + 14\). So the equation becomes:
\(3x+16=x + 14+2\sqrt{x + 13}\)
Subtract \(x + 14\) from both sides:
\(3x+16-(x + 14)=2\sqrt{x + 13}\)
\(2x+2 = 2\sqrt{x + 13}\)
Divide both sides by 2:
\(x + 1=\sqrt{x + 13}\)

Step4: Square both sides again

Square both sides to eliminate the remaining square root:
\((x + 1)^2=(\sqrt{x + 13})^2\)
\(x^2+2x + 1=x + 13\)

Step5: Solve the quadratic equation

Rearrange the equation to standard quadratic form \(ax^2+bx + c = 0\):
\(x^2+2x + 1-x - 13=0\)
\(x^2+x - 12=0\)
Factor the quadratic: We need two numbers that multiply to \(- 12\) and add to \(1\). The numbers are \(4\) and \(-3\). So,
\((x + 4)(x - 3)=0\)
Set each factor equal to zero:
\(x+4 = 0\) or \(x - 3=0\)
So \(x=-4\) or \(x = 3\)

Step6: Check for extraneous solutions

• For \(x=-4\):

Left - hand side of the original equation: \(\sqrt{3(-4)+16}-\sqrt{-4 + 13}=\sqrt{-12 + 16}-\sqrt{9}=\sqrt{4}-3=2 - 3=-1\)
Right - hand side: \(1\)
Since \(-1
eq1\), \(x = - 4\) is an extraneous solution.

• For \(x = 3\):

Left - hand side of the original equation: \(\sqrt{3(3)+16}-\sqrt{3 + 13}=\sqrt{9 + 16}-\sqrt{16}=\sqrt{25}-4=5 - 4 = 1\)
Right - hand side: \(1\)
Since \(1 = 1\), \(x = 3\) is a valid solution.

Answer:

\(3\)