# Solving Quadratic-Linear Systems

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

## Example 1

At the intersecting points,

both function's *y* values are the same.

So *x*^{2} - 2*x* = *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* = *x*^{2} - 2*x* and *y* = *x* + 4 are

(-1, 3) and (4, 8).

## Example 2

At the intersecting points,

both function's *y* values are the same.

So *x*^{2} - *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 *x*^{2} - 2*x* - *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)