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How do you find the discriminant of #x^2-2x+1=0# and use it to determine if the equation has one, two real or two imaginary roots?

Answer 1

Genesis Allen

#Delta =0 -># equation has a one real solution #x=1#

Consider the general form of the quadratic equation:

#ax^2+bx+c=0#

The discriminant #(Delta)# is defined as: #b^2-4ac#

Three cases arise:

(i) #Delta >0 -> # the equation has two distinct real roots (ii) #Delta <0 -> # the equation has two complex roots (iii) #Delta =0 -> # the equation has one real solution (Strictly, two equal real roots)

In our equation: #x^2-2x+1=0#

#Delta = (-2)^2-4*1*1 = 4-4 =0#

Hence the equation has one real solution.

The equation may be factorised as: #(x-1)(x-1)=0#

Hence the only real solution is #x=1#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.