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Can someone help me figure out and put on graphb)
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Can someone help me figure out and put on graphb)

Can someone help me figure out and put on graphb)-example-1
User Burcu
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Recall that:


\begin{gathered} For\text{ all }l\in\mathbb{R} \\ l^2\ge0. \end{gathered}

Notice that:


x^2-4x+7=x^2-4x+4+3=(x-2)^2+3.

Since x-2 is a real number, then:


(x-2)^2\ge0.

Therefore:


(x-2)^2+3\geq3>0.

Therefore the given equation has no real solutions.

Answer: The given equation has no real solutions.

User Chase Barker
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