The graphs of y=ax^2+bx+c and y=ax+b have one common point only, then?

Question

A. (a+b)^2-4ac=0
B. b^2-4ac=0
C. (a+b)^2+4ac=0
D. a^2+b^2-4ac=0
E. None of these

Answer 1

Answer is A

ax^2 + bx +c = ax +b

=> ax^2 +(b-a)x + c-b = 0

one common point means the above equation has a repeated root.
to have repeated root
(b-a)^2 - 4a(c-b) = 0
=> (a+b)^2 - 4ac = 0

Answer 2

All quadratics have 2 "roots", or factors. These can always be found by using the quadratic formula.
This says x={-b+or-rt(b^2-4ac)}/2a. a, b, are the coefficients of the x^2, and x terms respectively, while c is the constant
The (b^2-4ac) term in the formula can identify the nature of the roots.
If that term is positive, then a square root number exists and there are 2 roots: -b+ the rt and -b-the rt.
If b^2-4ac is 0' then x=-b. And the other root? It is -b also. Two real,equal roots,both the same. If you drew a graph, the quadratic would touch a straight line at the vertex.
If b^2-4ac is negative, then there are no real roots:you can't take the square root of a negative number.
To summarize:
b^2-4ac is positive, 2 roots ( cuts x-axis in 2 places)
b^2-4ac is zero, 2 equal roots (touches x-axis at vertex, but doesn't cut.
b^2-4ac is negative, no real roots. (doesn't cut or touch the x-axis)
The answer to your question is therefore B. b^2-4ac=0
And that's why the b^2-4ac term is called the "discriminant"

Answer 3

Set the equations equal to each other and combine:

ax^2+(b - a)x+(c - b) = 0

Solve for x:
x = (-(b - a) +/- SQRT((b - a)^2 - 4a(c - b))

The term in the SQRT must be = 0 for only one solution:
(b - a)^2 - 4a(c - b) = 0
b^2 - 2ab + a^2 - 4ac + 4ab = 0
b^2 + 2ab + a^2 - 4ac = 0
(b + a)^2 - 4ac = 0

so the answer is A