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descirbe the nature of the roots of then equation x^(2)-4x-x+4=0

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3 Answers

X2 - 4X - x +4 =0
 
X- 5X + 4 = 0
 
4 + 1 = 5      4(1) = 4
 
Therefore Quadratic can be factored into
 
 ( X - 4 ) ( X - 1 ) = 0
 
  X = 4  X = 1 are the roots, they are integers.  

This equation is x^2 - 4x - x + 4 = x^2 - 5x + 4 = 0. Factoring it yields (x - 1)(x - 4) = 0, so the roots are +1 and +4 which are both real. Another way to find the roots is to use the quadratic formula given by (-b + or - sqrt(b^2 - 4ac))/2a where a (the coefficient of the x^2 term) = 1, b (coefficient of the x term) = -5, and c (coefficient of the constant term) = 4. This gives (-(-5) + or - sqrt(5^2 - 4*1*4))/2*1 = (+5 + or - sqrt(25 - 16))/2 = (5 +- sqrt(9))/2 = (5 + or - 3)/2 = 8/2 (+4) or 2/2 (+1) as before. The answer is therefore b.

To identify the nature of roots, you have to analyze the discriminant (D)
 ax2 + bx + c = 0
D = b2 - 4ac
There are two real roots, if D > 0
There is one real root, if D = 0
There are two imaginary roots, if D < 0
~~~~~~~~~~~

x2 - 4x - x + 4 = 0
x2 - 5x + 4 = 0
a = 1, b = - 5, c = 4
D = (-5)2 - 4 • 1 • 4
D = 9 > 0
Thus, the answer is
b. Two real roots