Question

Can someone help me figure out and put on graphb)

Answer 1

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.