Question

solve.
\\(\sqrt{5x - 4} - \sqrt{x} - 5 = -3\\)

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

\\(\bigcirc\\) a. \\(x = \square\\)
\\(\quad\\) (type an integer or a simplified fraction. use a comma to separate answers as needed.)
\\(\bigcirc\\) b. there is no real solution.

Explanation:

Step1: Isolate one square root

First, we rewrite the equation \(\sqrt{5x - 4}-\sqrt{x}-5=-3\) as \(\sqrt{5x - 4}-\sqrt{x}=2\). Then, isolate \(\sqrt{5x - 4}\) to get \(\sqrt{5x - 4}=\sqrt{x}+2\).

Step2: Square both sides

Square both sides of the equation \(\sqrt{5x - 4}=\sqrt{x}+2\). Using the formula \((a + b)^2=a^{2}+2ab + b^{2}\), we have:
\[

$$\begin{align*} (\sqrt{5x - 4})^{2}&=(\sqrt{x}+2)^{2}\\ 5x-4&=x + 4\sqrt{x}+4 \end{align*}$$

\]

Step3: Simplify the equation

Subtract \(x\) and \(4\) from both sides:
\[

$$\begin{align*} 5x-4-x - 4&=4\sqrt{x}\\ 4x-8&=4\sqrt{x} \end{align*}$$

\]
Divide both sides by \(4\): \(x - 2=\sqrt{x}\)

Step4: Square both sides again

Square both sides of \(x - 2=\sqrt{x}\):
\[

$$\begin{align*} (x - 2)^{2}&=(\sqrt{x})^{2}\\ x^{2}-4x + 4&=x \end{align*}$$

\]
Rearrange to get a quadratic equation: \(x^{2}-5x + 4=0\)

Step5: Solve the quadratic equation

Factor the quadratic equation \(x^{2}-5x + 4=0\). We get \((x - 1)(x - 4)=0\), so \(x = 1\) or \(x = 4\)

Step6: Check for extraneous solutions

• For \(x = 1\):

Left - hand side of the original equation: \(\sqrt{5\times1-4}-\sqrt{1}-5=\sqrt{1}-1 - 5=1-1 - 5=-5
eq-3\)

• For \(x = 4\):

Left - hand side of the original equation: \(\sqrt{5\times4-4}-\sqrt{4}-5=\sqrt{16}-2 - 5=4-2 - 5=-3\)

So \(x = 4\) is a solution, but we made a mistake in step 3. Wait, let's re - check step 3.

Wait, going back to the equation \(\sqrt{5x - 4}-\sqrt{x}=2\). Let's consider the domain of the square roots:
For \(\sqrt{5x - 4}\), we need \(5x-4\geq0\Rightarrow x\geq\frac{4}{5}\)
For \(\sqrt{x}\), we need \(x\geq0\)
So the domain is \(x\geq\frac{4}{5}\)

Let's re - do step 2 and 3 correctly.

Starting from \(\sqrt{5x - 4}=\sqrt{x}+2\)

Squaring both sides: \(5x-4=x + 4\sqrt{x}+4\)

\(5x-4-x - 4=4\sqrt{x}\)

\(4x - 8=4\sqrt{x}\)

Divide by 4: \(x - 2=\sqrt{x}\)

Now, for \(x-2=\sqrt{x}\), we also need \(x - 2\geq0\) (because the left - hand side \(x - 2\) is equal to the square root \(\sqrt{x}\) which is non - negative), so \(x\geq2\)

Now, when \(x = 1\), \(x-2=-1\) and \(\sqrt{x}=1\), \(-1
eq1\), so \(x = 1\) is extraneous. When \(x = 4\), \(x - 2=2\) and \(\sqrt{x}=2\), so \(x = 4\) satisfies \(x-2=\sqrt{x}\)

Now, check \(x = 4\) in the original equation:

\(\sqrt{5\times4-4}-\sqrt{4}-5=\sqrt{16}-2 - 5=4 - 2-5=-3\), which matches the right - hand side.

Wait, but let's check the equation again. The original equation is \(\sqrt{5x - 4}-\sqrt{x}-5=-3\), which can be rewritten as \(\sqrt{5x - 4}-\sqrt{x}=2\)

When \(x = 4\): \(\sqrt{20 - 4}-\sqrt{4}=\sqrt{16}-2 = 4 - 2=2\), which is correct.

Wait, my initial check was wrong. The original equation is \(\sqrt{5x - 4}-\sqrt{x}-5=-3\), so \(\sqrt{5x - 4}-\sqrt{x}=2\). So when \(x = 4\), \(\sqrt{5\times4 - 4}-\sqrt{4}=4 - 2 = 2\), which is correct.

But let's consider the equation \(\sqrt{5x - 4}-\sqrt{x}-5=-3\)

Let's try to solve it again:

\(\sqrt{5x - 4}-\sqrt{x}=2\)

Let \(y=\sqrt{x}\), then \(x = y^{2}\) and the equation becomes \(\sqrt{5y^{2}-4}-y = 2\)

\(\sqrt{5y^{2}-4}=y + 2\)

Square both sides: \(5y^{2}-4=y^{2}+4y + 4\)

\(5y^{2}-y^{2}-4y-4 - 4=0\)

\(4y^{2}-4y - 8=0\)

Divide by 4: \(y^{2}-y - 2=0\)

Factor: \((y - 2)(y + 1)=0\)

So \(y = 2\) or \(y=-1\)

Since \(y=\sqrt{x}\geq0\), we discard \(y=-1\)

When \(y = 2\), \(\sqrt{x}=2\Rightarrow x = 4\)

Check \(x = 4\) in the original equation:

\(\sqrt{5\times4-4}-\sqrt{4}-5=\sqrt{16}-2 - 5=4 - 2-5=-3\), which is correct.

Answer:

A. \(x = 4\)