describe the nature of the roots of this equation 2x^2-x+1=0
x^2 - 2x + 1 = (x - 1)^2.
So, x^2 - 2x + 1 = 0
==> (x - 1)^2 = 0.
By the Zero-Product Property, x = 1 is the only solution. However, since the x - 1 factor is squared, 1 is a zero of multiplicity 2 and is thus a double-root.
Meaning, the answer is D.
Answer:
The given equation has two complex roots.
Step-by-step explanation:
The general form of a quadratic equation is
\(ax^2+bx+c=0\)
If \(b^2-4ac<0\), then the equation has two complex root.
If \(b^2-4ac=0\), then the equation has one real real root with multiplicity 2.
If \(b^2-4ac>0\), then the equation has two real root.
The given quadratic equation is
\(2x^2-x+1=0\)
Here a=2, b=-1 and c=1. The value of discriminant is
\(D=b^2-4ac=(-1)^2-4(2)(1)=1-8=-7\)
Since D<0, therefore the given equation has two complex roots.