Counter
Mathematics Dictionary
Dr. K. G. Shih
Completing the square
Subjects

Answers


Q01. Completing square in y = a*x^2 + b*x + c
Method
  • y = a*(x^2 + (b/a)*x + c/a).
  • y = a*(x^2 + (b/a)*x + c/a + (b/(2*a))^2 - (b/(2*a))^2 + c/a.
  • y = a*(x + b/(2*a))^2 - b^2/(4*a) + c.
  • y = a*(x + b/(2*a))^2 - (b^2 - 4*a*c)/(4*a).
Application 1 : Find vertex
  • At vertex of y = a*x^2 + b*x + c, it has a minimum or a maximam.
  • Hence (x + b/(2*a) = 0 at vertex.
  • Vertex : xv = -b/(2*a) and yv = F(xv)
Application 2 : Quadratic formula of a*x^2 + b*x + c = 0
  • Since y = 0 in y = a*x^2 + b*x + c.
  • Hence (x + b/(2*a)^2 = (b^2 - 4*a*c)/(4*a).
  • Hence x = (-b + Sqr(b^2 - 4*a*c))/(2*a).
  • Hence x = (-b - Sqr(b^2 - 4*a*c))/(2*a).
Example : Solve x^2 - 6*x + 8 = 0 by quadratic formula
  • Since a = 1, b = -6 and c = 8, hence
  • x1 = (-b + Sqr(b^2 - 4*a*c))/(2*a)
  • x1 = (-(-6) + Sqr(36 - 4*1*8))/(2*1)
  • x1 = (6 + Sqr(4))/2 = 4.
  • x2 = (-b - Sqr(b^2 - 4*a*c))/(2*a)
  • x2 = (-(-6) - Sqr(36 - 4*1*8))/(2*1)
  • x2 = (6 - Sqr(4))/2 = 2.
Go to Begin

Q02. Completing the square in circle equation
Circle standard equation : (x - h)^2 + (y - k)^2 = r^2
  • Center is at (h,k).
  • Radius is r.
Circle equation is x^2 + y^2 + c*x + d*y + e = 0, find center and radius.
  • (x^2 + c*x + (c/2)^2 - (c/2)^2) + (y^2 + d*x + (d/2)^2 - (d/2)^2) + e = 0.
  • (x + c/2)^2 + (y + d/2)^2 = (c/2)^2 + (d/2)^2) - e.
  • Hence center is at (-c/2,-d/2) and radius is r = Sqr((c/2)^2 + (d/2)^2) - e).
  • Note 1 : If r > 0 it is a circle.
  • Note 2 : if r = 0 it is a point.
  • Note 3 : if r < 0 it does not exist in the real number system.
Go to Begin

Q03. Completing the square in ellipse equation
Ellipse standard equation : (x - h)^2/(A^2) + (y - k)^2/(B^2) = 1
  • Center is at (h,k).
  • Semi-axese are A and B.
Circle equation is a*x^2 + b*y^2 + c*x + d*y + e = 0, find center and semi-axese.
  • a*(x^2 + c*x/a + (c/(2*a))^2 - (c/(2*a))^2)
  • a*(x + c/(2*a))^2 + b*(y + d/(2*b))^2 = (c/(2*a))^2 + (d/(2*b))^2) - e.
  • Let R = Sqr((c/(2*a))^2 + (d/(2*b))^2) - e).
  • a*(x + c/(2*a))^2/R + b*(y + d/(2*b))^2/R = 1.
  • Hence center is at (-c/(2*a),-d/(2*b)).
  • Semi axese : A = R/a and B = R/b.
  • Note 1 : If R > 0 it is an ellipse.
  • Note 2 : if R = 0 it is a point.
  • Note 3 : if R < 0 it does not exist in the real number system.
Go to Begin

Q04. Completing the square in hyperbola equation
Hyperbola standard equation : (x - h)^2/(A^2) - (y - k)^2/(B^2) = 1
  • Center is at (h,k).
  • Semi-axese are A and B.
Circle equation is a*x^2 - b*y^2 + c*x + d*y + e = 0, find center and semi-axese.
  • a*(x^2 + c*x/a + (c/(2*a))^2 - (c/(2*a))^2)
  • a*(x + c/(2*a))^2 - b*(y + d/(2*b))^2 = (c/(2*a))^2 - (d/(2*b))^2) - e.
  • Let R = Sqr((c/(2*a))^2 - (d/(2*b))^2) - e).
  • a*(x + c/(2*a))^2/R - b*(y + d/(2*b))^2/R = 1.
  • Hence center is at (-c/(2*a),-d/(2*b)).
  • Semi axese : A = R/a and B = R/b.
  • Note 1 : If R > 0 it is a hyperbola.
  • Note 2 : if R = 0 it is a point.
  • Note 3 : if R < 0 it does not exist in the real number system.
Go to Begin

Q05. Reference
 Go to Begin

Q06. Answer
Go to Begin

Q07. Answer
 Go to Begin

Q08. Answer
 Go to Begin

Q09. Answer
 Go to Begin

Q10. Answer
 Go to Begin
Show Room of MD2002 Contact Dr. Shih Math Examples Room
Copyright © Dr. K. G. Shih, Nova Scotia, Canada.
1