Answered

Westonci.ca is the premier destination for reliable answers to your questions, brought to you by a community of experts. Join our platform to connect with experts ready to provide precise answers to your questions in different areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.

How would you find the real solution, by using the quadratic formula.
x^2-4x+2=0

Sagot :

naǫ
[tex]x^2-4x+2=0 \\ \\ a=1 \\ b=-4 \\ c=2 \\ b^2-4ac=(-4)^2-4 \times 1 \times 2=16-8=8 \\ \\ x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}=\frac{-(-4) \pm \sqrt{8}}{2 \times 1}=\frac{4 \pm \sqrt{4 \times 2}}{2}=\frac{4 \pm 2\sqrt{2}}{2}=\frac{2(2 \pm \sqrt{2})}{2}=2 \pm \sqrt{2} \\ \boxed{x=2-\sqrt{2} \hbox{ or } x=2+\sqrt{2}}[/tex]
konrad509
[tex]x^2-4x+2=0\\x^2-4x+4-2=0\\ (x-2)^2=2\\ x-2=\sqrt2 \vee x-2=-\sqrt2\\ x=2+\sqrt2 \vee x=2-\sqrt2[/tex]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.