How do you solve #x^2- x + 41# using the quadratic formula?

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Answer 1 · 2018-07-19T22:26:42

#x=(1+isqrt163)/2# and #x=(1-isqrt163)/2#

We can find the roots of any quadratic of the form #ax^2+bx+c# with the Quadratic Formula

#bar( ul|color(white)(2/2)x=(-b+-sqrt(b^2-4ac))/(2a)color(white)(2/2)|)#

We have the quadratic #x^2-x+41#, where #a=1, b=-1# and #c=41#. Plugging these values in, we get

#x=(1pmsqrt(1-4(1*41)))/2#

This simplifies to

#x=(1pmsqrt(-163))/2# or

#x=1/2 pm sqrt(-163)/2#

We can rewrite #sqrt(-163)# as #sqrt(163)*sqrt(-1)#, or #isqrt3#. Doing this, we now have

#x=(1+isqrt163)/2# and #x=(1-isqrt163)/2#

Hope this helps!