Question

which of the following describes the solution to the equation \\(\sqrt{c^3 + 2c - 4} = -\sqrt{1 - 2c}\\)?

• \\(-5\\) is an extraneous solution, and (1) is a true solution.
• \\(-5\\) is a true solution, and (1) is an extraneous solution.
• both \\(-5\\) and (1) are true solutions.
• both \\(-5\\) and (1) are extraneous solutions.

Explanation:

Step1: Square both sides

To solve the equation \(\sqrt{c^{2}+2c - 4}=-\sqrt{1 - 2c}\), we first square both sides to eliminate the square roots. Squaring both sides gives us \((\sqrt{c^{2}+2c - 4})^{2}=(-\sqrt{1 - 2c})^{2}\), which simplifies to \(c^{2}+2c - 4=1 - 2c\).

Step2: Rearrange into quadratic equation

Rearrange the equation \(c^{2}+2c - 4=1 - 2c\) to standard quadratic form \(ax^{2}+bx + c = 0\). We get \(c^{2}+2c+2c - 4 - 1=0\), which simplifies to \(c^{2}+4c - 5=0\).

Step3: Factor the quadratic equation

Factor the quadratic equation \(c^{2}+4c - 5=0\). We need two numbers that multiply to \(- 5\) and add to \(4\). The numbers are \(5\) and \(-1\), so the factored form is \((c + 5)(c - 1)=0\).

Step4: Solve for c

Set each factor equal to zero: \(c+5 = 0\) or \(c - 1=0\). Solving these gives \(c=-5\) or \(c = 1\).

Step5: Check for extraneous solutions

We need to check if these solutions satisfy the original equation, especially considering the domain of square roots (the expression inside the square root must be non - negative) and the sign of the square root (the square root of a non - negative number is non - negative, and the right - hand side of the original equation is \(-\sqrt{1 - 2c}\), so the right - hand side must be non - positive).

Check \(c = 1\):

For the right - hand side, when \(c = 1\), \(-\sqrt{1-2(1)}=-\sqrt{-1}\), which is not a real number. So \(c = 1\) is an extraneous solution.

Check \(c=-5\):

For the left - hand side, \(\sqrt{(-5)^{2}+2(-5)-4}=\sqrt{25-10 - 4}=\sqrt{11}\). For the right - hand side, \(-\sqrt{1-2(-5)}=-\sqrt{1 + 10}=-\sqrt{11}\). But \(\sqrt{11}
eq-\sqrt{11}\) (since the left - hand side of the original square root equation \(\sqrt{c^{2}+2c - 4}\) is non - negative and the right - hand side \(-\sqrt{1 - 2c}\) is non - positive, and they can only be equal if both are zero. But \(\sqrt{11}\) and \(-\sqrt{11}\) are not zero and are negatives of each other). Wait, let's re - check the domain for \(c=-5\):

For the left - hand side: \(c^{2}+2c - 4=(-5)^{2}+2\times(-5)-4=25 - 10 - 4 = 11\geq0\)

For the right - hand side: \(1-2c=1-2\times(-5)=1 + 10 = 11\geq0\), but the left - hand side \(\sqrt{11}\) is positive and the right - hand side \(-\sqrt{11}\) is negative. So \(c=-5\) also does not satisfy the original equation. Wait, there is a mistake in the previous check. Let's re - evaluate:

The original equation is \(\sqrt{c^{2}+2c - 4}=-\sqrt{1 - 2c}\). The left - hand side \(\sqrt{c^{2}+2c - 4}\geq0\) and the right - hand side \(-\sqrt{1 - 2c}\leq0\). For them to be equal, both sides must be zero.

Set \(\sqrt{c^{2}+2c - 4}=0\) and \(-\sqrt{1 - 2c}=0\).

From \(\sqrt{c^{2}+2c - 4}=0\), we have \(c^{2}+2c - 4 = 0\), and from \(-\sqrt{1 - 2c}=0\), we have \(1-2c = 0\) or \(c=\frac{1}{2}\). But \(c=\frac{1}{2}\) does not satisfy \(c^{2}+2c - 4=0\) (\((\frac{1}{2})^{2}+2\times\frac{1}{2}-4=\frac{1}{4}+1 - 4=\frac{1 + 4-16}{4}=-\frac{11}{4}
eq0\)).

When we solved the equation by squaring, we introduced an extraneous solution. In fact, when we check \(c=-5\):

Left - hand side: \(\sqrt{(-5)^{2}+2\times(-5)-4}=\sqrt{25 - 10-4}=\sqrt{11}\)

Right - hand side: \(-\sqrt{1-2\times(-5)}=-\sqrt{11}\)

Since \(\sqrt{11}
eq-\sqrt{11}\), \(c = - 5\) does not satisfy the original equation. When we check \(c = 1\):

Left - hand side: \(\sqrt{1^{2}+2\times1-4}=\sqrt{1 + 2-4}=\sqrt{-1}\) (not a real number)

So both \(c=-5\) and \(c = 1\) are extraneous solutions.

Answer:

Both \(-5\) and \(1\) are extraneous solutions.