How do you solve #x^2-10x+41=0#?

1 Answer
Y
May 11, 2017

Original question: Solve #x^2-10x+41=0#

Since the equation cannot be easily factored, we must use the quadratic equation:
#x=(-b+-sqrt(b^2-4ac))/(2a)# where #a,b,c# are from the standard form of a quadratic, #f(x)=ax^2+bx+c#

In this question, #a=1,b=-10,c=41#, which we can substitute into the quadratic equation and simplify:
#x=(-(-10)+-sqrt((-10)^2-4*1*41))/(2*1)#
#x=(10+-sqrt(100-164))/2#
#x=(10+-sqrt(-64))/2#

Since we cannot square root a negative number, we use #i# to denote the imaginary unit, #i=sqrt(-1)#, and continue simplifying:
#x=(10+-8i)/2#
#x=5+-4i#

Therefore, our solutions are:
#x=5+4i,5-4i#

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