Is it possible to factor $x^2-6x+7$ over $\mathbb{R}$
None of the online calculators seem to give me an answer.
I am trying to find the values for x. How do I do this again?
$$x^2-6x = -7$$
Then what?
$\endgroup$ 13 Answers
$\begingroup$$$ x^2-6x+7=(x^2-6x+9)-2=(x-3)^2-(\sqrt{2})^2=(x-3-\sqrt{2})(x-3+\sqrt{2}). $$
$\endgroup$ 2 $\begingroup$Yes, it is possible, you know this since the discriminant is greater than $0$:$$\Delta=b^2-4ac=36-4\cdot 1\cdot 7=8,$$being the coefficients $a=1$, $b=-6$, $c=7$. Follows that $\sqrt{\Delta}=2\sqrt{2}$ and$$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=3\pm\sqrt{2}.$$Now,$$ \begin{align} x^2-6x+7&=(x-x_1)(x-x_2)=\\ &=(x-(3+\sqrt{2}))(x-(3-\sqrt{2}))=\\ &=(x-3-\sqrt{2})(x-3+\sqrt{2}).\\ \end{align} $$
$\endgroup$ $\begingroup$You can factor it over $\mathbb R$ as $(x-3-\sqrt2)(x-3+\sqrt2)$!
$\endgroup$
