Question

what is the solution to the equation \\(\sqrt{4t + 5} = 3 - \sqrt{t + 5}\\)?
\\(\bigcirc\\) \\(x = -1, x = 11\\)
\\(\bigcirc\\) \\(x = -1\\)
\\(\bigcirc\\) \\(x = 11\\)
\\(\bigcirc\\) no solution

Explanation:

Step1: Isolate a square root

Start with the equation \(\sqrt{4t + 5}=3-\sqrt{t + 5}\). Let's isolate one of the square roots. Let's move \(-\sqrt{t + 5}\) to the left side: \(\sqrt{4t + 5}+\sqrt{t + 5}=3\). Wait, actually, maybe it's better to first square both sides to eliminate the square roots. Let's square both sides: \((\sqrt{4t + 5})^2=(3 - \sqrt{t + 5})^2\).
Simplify both sides: \(4t + 5 = 9-6\sqrt{t + 5}+(t + 5)\).

Step2: Simplify the equation

Simplify the right side: \(4t + 5 = t + 14-6\sqrt{t + 5}\).
Subtract \(t + 14\) from both sides: \(4t + 5-(t + 14)=-6\sqrt{t + 5}\).
Simplify left side: \(3t - 9=-6\sqrt{t + 5}\).
Divide both sides by 3: \(t - 3=-2\sqrt{t + 5}\).

Step3: Square both sides again

Now square both sides to eliminate the remaining square root: \((t - 3)^2=(-2\sqrt{t + 5})^2\).
Simplify both sides: \(t^2-6t + 9 = 4(t + 5)\).
Expand right side: \(t^2-6t + 9 = 4t + 20\).

Step4: Solve the quadratic equation

Bring all terms to left side: \(t^2-6t + 9-4t - 20 = 0\).
Simplify: \(t^2-10t - 11 = 0\).
Factor the quadratic: \((t - 11)(t + 1)=0\).
So the solutions are \(t = 11\) or \(t=-1\). But we need to check for extraneous solutions because we squared both sides (which can introduce extraneous solutions).

Step5: Check solutions

First, check \(t = 11\):
Left side: \(\sqrt{4(11)+5}=\sqrt{44 + 5}=\sqrt{49}=7\).
Right side: \(3-\sqrt{11 + 5}=3-\sqrt{16}=3 - 4=-1\).
But \(7
eq - 1\), so \(t = 11\) is extraneous.
Now check \(t=-1\):
Left side: \(\sqrt{4(-1)+5}=\sqrt{-4 + 5}=\sqrt{1}=1\).
Right side: \(3-\sqrt{-1 + 5}=3-\sqrt{4}=3 - 2 = 1\).
So \(t=-1\) is a valid solution. Wait, but let's check my steps again. Wait, when I squared the first time, maybe I made a mistake. Wait, original equation: \(\sqrt{4t + 5}=3-\sqrt{t + 5}\). Let's plug \(t=-1\): left side \(\sqrt{4(-1)+5}=\sqrt{1}=1\), right side \(3-\sqrt{-1 + 5}=3 - 2 = 1\). Good. For \(t = 11\): left side \(\sqrt{4(11)+5}=\sqrt{49}=7\), right side \(3-\sqrt{11 + 5}=3 - 4=-1\). 7≠-1, so \(t = 11\) is extraneous. So only \(t=-1\) is valid. Wait, but the options are in terms of \(x\), maybe it's a typo and the variable is \(x\) instead of \(t\). So the solution is \(x=-1\). Wait, but let's check the options. The options are \(x=-1,x = 11\); \(x=-1\); \(x = 11\); no solution. So after checking, only \(x=-1\) is valid.

Answer:

B. \(x=-1\) (assuming the option is labeled as B, but in the given options, the second option is \(x=-1\))