Steply

Solve y = x^2 + 3 and y = x

Step by step, with the reason for every step · Nonlinear systems · Algebra 2

Step by step

The question

Question: solve y = x² + 3 and y = x.

y = x2 + 3①

We call them ① and ②.

We want the points (x, y) that make both true.

y = x②

① is a parabola, and ② is a line.

Not both lines, so this is called a nonlinear system.

y = ax² + ...: a parabola
y = mx + b: a line

1 Put ② into ①

In ②, y is already alone: y = x.

Both ① and ② say what y is.

So the two right sides are equal.

x2 + 3 = x

2 Get 0 on one side

We take x away from both sides.

x2 + 3 − x = x − x

Careful!

The answers are points, not just x numbers.

After x, we must find each y.

Careful: each x needs its y: the answers are points

Now we have x² − x + 3 = 0.

x2 − x + 3 = 0

3 Can it be 0?

It does not factor.

b² − 4ac, the part under the root in the formula, is −11.

Negative.

b² − 4ac = (−1)² − 4 × 1 × 3 = −11

No real number has a negative square root.

So no x works: the line and the parabola never meet.

There is no solution.

= no real x: no solution

We can see it on a picture.

We draw both on a grid.

① is a parabola.

② is a line.

Let's look back.

First: put ② into ①.

Then: get 0 on one side.

Last: can it be 0?.

Answer
= no real x: no solution

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