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which equation results from isolating a radical term and squaring both …

Question

which equation results from isolating a radical term and squaring both sides of the equation for the equation (sqrt{c - 2}-sqrt{c}=5)?

(c - 2 = 25 + c)

(c - 2 = 25 - c)

(c - 2 = 25 + c - 10sqrt{c})

(c - 2 = 25 + c + 10sqrt{c})

Explanation:

Step1: Isolate the radical term

Given the equation \(\sqrt{c - 2}-\sqrt{c}=5\), we first isolate one of the radical terms. Let's isolate \(\sqrt{c - 2}\) by adding \(\sqrt{c}\) to both sides of the equation.
We get \(\sqrt{c - 2}=5+\sqrt{c}\).

Step2: Square both sides of the equation

Now, we square both sides of the equation \(\sqrt{c - 2}=5+\sqrt{c}\).
Recall that \((a + b)^2=a^{2}+2ab + b^{2}\). Here, \(a = 5\) and \(b=\sqrt{c}\). So, \((\sqrt{c - 2})^{2}=(5+\sqrt{c})^{2}\).
Simplifying the left - hand side, \((\sqrt{c - 2})^{2}=c - 2\).
Simplifying the right - hand side, \((5+\sqrt{c})^{2}=5^{2}+2\times5\times\sqrt{c}+(\sqrt{c})^{2}=25 + 10\sqrt{c}+c\).
So, after squaring both sides, we have \(c - 2=25 + c+10\sqrt{c}\).

Answer:

\(c - 2=25 + c+10\sqrt{c}\) (the fourth option: \(c - 2=25 + c+10\sqrt{c}\))