If ax^2 + bx + c = a (x −d)^2 with a ̸= 0, find d in terms of a, b and c.

Answer 1

IF ax^2 + bx + c = a(x-d)^2

then the graph is a parabola with vertex & local&absolute extremum at (d,0)

a(x^2 -2d+d^2)=

ax^2-2adx +ad^2

set the coefficient of the x terms equal

and set the constant terms equal

set the coefficients equal

b=-2ad

d = -b/2a (1st part of the quadratic formula: x=-b/2a +/-(1/2a)sqr(a^2 -4ac). 1st part, -b/2a = x coordinate of the vertex )

set constant terms equal

c=ad^2

d^2 = c/a

d = +/-sqr(c/a)

d= -b/2a and +/-sqr(c/a)

-b/2a = +/-sqr(c/a)

c/a = b^2/4a^2

c = b^2/4a

b^2 = 4ac

b=+/-2sqr(ac)

discriminant = b^2 -4ac = 4ac-4ac = 0

x =-b/2a = d= the only zero= the only x "intercept" (the parabola is tangent to the x axis and doesn't cross it) = the only solution for x

x = -b/2a +/-sqr(0) = -b/2a, just one solution, not the usual two

a never = 0, as then the parabola would degenerate into a straight line, a linear equation, no longer a quadratic