Please solve the following quadratic equation and show your work: x^2-2x+6=0

1 answer:

6 0

$ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\if\ \Delta > 0\ then\ exist\ two\ solutions\\\\x_1=\dfrac{-b-\sqrt\Delta}{2a}\ and\ x_2=\dfrac{-b+\sqrt\Delta}{2a}\\\\if\ \Delta =0\ then\ exist\ one\ solution\ x_o=\dfrac{-b}{2a}\\\\if\ \Delta < 0\ then\ no\ solutions$

We have:

$x^2-2x+6=0\\\\a=1;\ b=-2;\ c=6\\\\\Delta=(-2)^2-4\cdot1\cdot6=4-24=-20 < 0$

Answer: NO SOLUTIONS.

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Notice the picture below

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$\bf \textit{parabola vertex form with focus point distance}\\\\
\begin{array}{llll}
(y-{{ k}})^2=4{{ p}}(x-{{ h}}) \\\\
\boxed{(x-{{ h}})^2=4{{ p}}(y-{{ k}}) }\\
\end{array}
\qquad 
\begin{array}{llll}
vertex\ ({{ h}},{{ k}})\\\\
{{ p}}=\textit{distance from vertex to }\\
\qquad \textit{ focus or directrix}
\end{array}\\\\
-----------------------------\\\\$

$\bf \begin{cases}
p=-1\\
h=-2\\
k=5
\end{cases}\implies (x-(-2))^2=4(-1)(y-5)
\\\\\\
(x+2)^2=-4(y-5)\implies 
-\cfrac{1}{4}(x+2)^2=y-5
\\\\\\
\boxed{-\cfrac{1}{4}(x+2)^2+5=y}$