Question

using the method of completing the square, rewrite $x^{2}-4x + 7$ in the form $(x + b)^{2}+c$.
$x^{2}-4x + 7=$

Explanation:

Step1: Isolate the $x$ - terms

$x^{2}-4x + 7=(x^{2}-4x)+7$

Step2: Complete the square for the $x^{2}-4x$ part

For the quadratic expression $x^{2}-4x$, the coefficient of $x$ is $-4$. Half of it is $\frac{-4}{2}=-2$, and its square is $(-2)^{2} = 4$. We add and subtract 4 inside the parentheses:
$(x^{2}-4x)+7=(x^{2}-4x + 4-4)+7$

Step3: Rewrite the perfect - square trinomial

$x^{2}-4x + 4=(x - 2)^{2}$, so $(x^{2}-4x + 4-4)+7=(x - 2)^{2}-4 + 7$

Step4: Simplify the constant terms

$(x - 2)^{2}-4 + 7=(x - 2)^{2}+3$

Answer:

$(x - 2)^{2}+3$