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Solving a Quadratic Equation

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Solving Quadratic Equation

1. The equation is of the form a(x − h)^2 + k = 0
Corresponding parabola or quadratic function: y = a(x − h)^2 + k
Solutions are x-intercepts of this parabola

don’t forget ± sign

• Solutions are

• Simplify and write as 2 separate numbers if − k/a is a perfect square

2. The equation is not in the above form.

• If the equation is not in the form ax^2 + bx + c = 0, then bring every term on one
side of “=”, foil (if necessary) and simplify to ax^2 + bx + c = 0
Corresponding parabola or quadratic function: y = ax^2 + bx + c
Solutions are x-intercepts of this parabola

• The solution is

Simplify and write as 2 separate numbers if b^2 − 4ac is a perfect square

You will get:

Discriminant

Type of solution

(* if p, q, r or a, b, c are integers)

Graphically
positive

perfect square
not a perfect square

2 real solutions

2 rational solution*
2 real solution with radicals
conjugate to each other

2 x-intercepts

parabola crosses x-axis twice

zero 1 rational solution* only 1 x-intercept
parabola just touches x axis
negative 2 complex solutions
conjugate to each other
no x-intercept
parabola does not intersect x axis

Graphing Quadratic Function: Vertical Parabola

1. The function is in the form y = a(x − h)^2 + k
• Plot the points (h, k), (h + 1, k + a) and (h − 1, k + a)
• Draw the parabola through these points.

2. The function is in the form y = ax^2 + bx + c

• Plot the points:
y-intercept - (0, c), Point of symmetry(or
plug in x = − b/2a into y = ax^2 + bx + c to get y-coordinate of vertex)

• Draw the parabola through these points.

• a > 0 - parabola opens up (smilie) with minimum at the vertex
• a < 0 - parabola opens down (frownie) with maximum at the vertex

  y = a(x − h)^2 + k y = ax^2 + bx + c
vertex (h, k)
axis x = h
symmetric points* (h + 1, k + a) and (h − 1, k + a) (0, c) and
y-intercept (0, ah^2 + k) (0, c)
x-intercept    
none if k > 0 if b^2 < 4ac
one if k = 0 then (h, 0) if b^2 = 4ac then
two if k < 0 then if b^2 > 4ac then

* These points are on opposite sides of the axis, at equal distance from the axis and are at
the same height i.e. they have the same y-coordinate.

For horizontal parabola:

x = a(y − k)^2 + h - plot (h, k), (h + a, k + 1) and (h + a, k − 1) and draw the parabola
x = ay^2 + by + c - plot (c, 0),and draw the parabola
a > 0 - parabola opens right, a < 0 - parabola opens left