How to solve the quadratic-linear system and use the discriminant to find the number of the roots: examples and their solutions.

Example 1: Solve y = x2 - 2x, y = x + 4

Both equations have the same y.

So x2 - 2x = x + 4.

Move x + 4 to the left side.

Then x2 - 3x - 4 = 0.

Factor x2 - 3x - 4.

Find a pair of numbers
whose product is the constant term [-4]
and whose sum is the middle term's coefficient [-3].

The constant term is (-).
So the signs of the numbers are different:
one is (+), and the other is (-).

(-1, 4) are not the right numbers.

[-4] = -4⋅1
-4 + 1 = [-3]
So -4 and 1 are the right numbers.

Use -4 and +1
to write a factored form:
(x - 4)(x + 1) = 0.

Solve (x - 4)(x + 1) = 0.

1) x - 4 = 0
So x = 4.

Solving a quadratic equation by factoring

x = 4

Put this into [y = x + 4].
(You can also put this into [y = x2 - 2x].)

Then y = 8.

Substitution method

x = 4
y = 8

So, for case 1,
(x, y) = (4, 8).

See case 2 of (x - 4)(x + 1) = 0.

2) x + 1 = 0
So x = -1.

x = -1

Put this into [y = x + 4].

Then y = 3.

x = -1
y = 3

So, for case 2,
(x, y) = (-1, 3).

So (x, y) = (-1, 3), (4, 8).

Let's see what the answer means.

Below are the graphs of the system:

y = x2 - 2x
y = x + 4.

As you can see,
the solution of the system
are the intersecting points of the functions:
(-1, 3), (4, 8).

Example 2

Both functions have the same y.

So x2 - x = x + k.

Move x + k to the left side.

Then x2 - 2x - k = 0.

Recall that
the discriminant of a quaratic equation
determines the number of the roots.

The discriminant

If the given functions intersect,
there would be at least one intersecting point.

Then there would be at least one x.

So set D ≥ 0.

a = 1
b = -2
c = -k

Then D = (-2)2 - 4⋅1⋅(-k).
And this is greater than or equal to 0.

(-2)2 = 4
-4⋅1⋅(-k) = +4k

Move 4 to the right side.

Then 4k ≥ -4.

Divide both sides by 4.

Then k ≥ -1.

Let's see the relationship
between the discriminant D
and the number of the intersecting points.

Below are the graphs of the given functions:

y = x2 - x
y = x + k.

If D > 0,
there are two roots.

So there are two intersecting points.

If D = 0,
there's one root.

So there's one intersecting point:
[y = x + k] touches [y = x2 - x].

If D < 0,
there's no root.

So there's no intersecting point.