How do you solve x^2-10x+41=0x2−10x+41=0?

1 Answer
Y
May 11, 2017

Original question: Solve x^2-10x+41=0x2−10x+41=0

Since the equation cannot be easily factored, we must use the quadratic equation:
x=(-b+-sqrt(b^2-4ac))/(2a)x=−b±√b2−4ac2a where a,b,ca,b,c are from the standard form of a quadratic, f(x)=ax^2+bx+cf(x)=ax2+bx+c

In this question, a=1,b=-10,c=41a=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)±√(−10)2−4⋅1⋅412⋅1
x=(10+-sqrt(100-164))/2x=10±√100−1642
x=(10+-sqrt(-64))/2x=10±√−642

Since we cannot square root a negative number, we use ii to denote the imaginary unit, i=sqrt(-1)i=√−1, and continue simplifying:
x=(10+-8i)/2x=10±8i2
x=5+-4ix=5±4i

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

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