
Mike C. answered 08/05/13
Enthusiastic Tutor for Middle and High School Students
x2 - 2x + 3 = (x - 1)2 + 2
= (x - 1)2 - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
Timothy S.
asked 08/05/13answer for math questions about parabolas
Mike C. answered 08/05/13
Enthusiastic Tutor for Middle and High School Students
x2 - 2x + 3 = (x - 1)2 + 2
= (x - 1)2 - (-2)
= (x - 1 + sqrt(-2)) (x - 1 - sqrt(-2))
= (x - 1 + sqrt(2) * i) (x - 1 - sqrt(2) * i)
Stuart R. answered 08/05/13
Math Tutor - Online and In Home -20 Years Experience
Parabolas are of the form ax2+bx+c, where a,v and c are the coefficients (numbers) in front of the variable. Notice that all parabolas are of degree 2 (that means the highest power of the variable is 2).
There are three major methods of solving for the roots of a parabola. (roots being the places where the y value will be zero).
1. factor the tri-nomial (three terms) into two binomials (two terms).
2. complete the square (see below)
3. or use the quadratic formula -b + sqrt(b2 - 4ac) then divide the whole thing by 2a
for the parabola stated above: y =
x2 - 2x + 3 factoing will not be the best way.
completing the square is done as follows:
set the equation = 0. 0 = x2 -2x + 3
take the last term (3) and subtract it from both sides which gives you x2 - 2x = -3
take the coefficient of the 'x' term which is -2, take half of it (-1) square it (1)
add this term (1) to both sides which gives you x2 - 2x +1 = -2
NOTICE: you have made a perfect square of the left side which is (x-1)2 so....
(x-1)2 =-2. Now take the sqrt of both sides which gives you x-1 =+sqrt(-2)
so..... the roots are complex! not real. they are x = 1-i*sqrt(2) and 1+i*sqrt(2)
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NOTE: The vertex or (maximum/minimum point) of the parabola ca be gotten by using the co-ordinate pair (-b/2a,f(-b/2a))
((-2/2),6) or ((-1,6)
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Using the quadratic formula we get -(-2) + sqrt(-22-4(a)(3) all divided by (2*1)
which gives you 2 + sqrt(4 - 12) all divided by 2
which gives you 2 + sqrt(-8) all divided by 2, which yields 2 + 2i* sqrt(2) alld divided by 2
whihc is 1 + i*sqrt(2) and 1 - i*sqrt(2)
same answer as completing the square!
Grigori S. answered 08/05/13
Certified Physics and Math Teacher G.S.
Where are the qustions? Your parabola is opened up because the first coefficient is positive, The parabola has two roots defined by the equation
x2 -2x + 3 = 0
Factor it:
x2 -2x +3 = (x-3)(x+1) = 0
Thus the roots are x1 = 3 and x2 = -1. The axis of symmetry for the parabola is the midpoint of the segment [-1,3], which means x - coordinate of the vertex is -1 + [3-(-1)]/2 = 1. Because the parabola is opened up it has only minimum which is equal to y(1) = 2.
Grigori S.
You are right. I messed up with the sign, being in hurry and didn't pay attention on it. The equation doesn't have a real solutions. Quadratic formula gives us:
x1 = 1 + iv2, x2 = 1 - iv2
because the discriminanat is negative. There is no intersection with x -axis. The line of symmetry for the parabola is still the same: x = -b/2a, where b = -2 and a = 1. I apologize for accidental inaccuracy. Thank you for pointing to my miscalcultions.
08/06/13
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Mike C.
Wrong solution Grigori. Your factoring works for x2 - 2x - 3.
08/05/13