Solving Quadratic-Linear Systems

Solving Quadratic-Linear Systems

How to solve the quadratic-linear system problems: examples and their solutions.

Example 1

Solve the system of equations. y = x^2 - 2x, y = x + 4

At the intersecting points,
both function's y values are the same.

So x2 - 2x = x + 4.

Substitution method

Then x = -1, 4.
These are the intersecting points' x values.

Solving a quadratic equation by factoring

Put x = -1, 4
into y = x + 4.

Then y = 3, 8.

So the intersecting points are (-1, 3) and (4, 8).

You can see that
the intersecting points of
y = x2 - 2x and y = x + 4 are
(-1, 3) and (4, 8).

Example 2

Find the range of k that makes the given functions intersect. y = x^2 - x, y = x + k

At the intersecting points,
both function's y values are the same.

So x2 - x = x + k.

Substitution method

Recall that
the discriminant of a quadratic equation
determines the nature of the roots:

If D ≥ 0,
then there are 1 or 2 real roots.

So, for x2 - 2x - k = 0,
if D ≥ 0,
then there are 1 or 2 intersecting points.
(= The functions intersect.)

So k ≥ -1.

If D > 0 (k > -1),
then there are two intersecting points.

If D = 0 (k = -1),
then there's one intersecting point.

If D < 0 (k < -1),
then there are no intersecting points.
(no real roots)