Quadratic Systems in Two Unknowns

What this move does

An equation in two unknowns of degree at most two has a curve as its graph: a circle, a parabola, a hyperbola, or simply a straight line. The solutions of the system are the points the two curves have in common.

The most common kind: one equation is linear and the other is quadratic. The method is the same as for a system of two linear equations, elimination by substitution:

{2 x - y = 5 x^{2} + y^{2} = 10 \Rightarrow x^{2} + (2 x - 5)^{2} = 10

Use the linear equation to write one unknown in terms of the other.
Substitute into the quadratic equation to get a quadratic equation in one unknown. Its discriminant $Δ$ decides how many solutions there are.
Solve for that unknown, then substitute back into the linear equation to find the other one.

A line and a quadratic curve have at most two points in common, so a system like this has at most two solutions: two if $Δ > 0$ , one if $Δ = 0$ (tangent), and none if $Δ < 0$ .

Work through one

Example: solve 2x − y = 5, x² + y² = 10

Step 1 / 6

Look at the structure
① is a linear equation, whose graph is a line; ② is quadratic, and its graph is a circle. We are looking for the points the line and the circle share.

${2 x - y = 5 x^{2} + y^{2} = 10 ① ②$
Solve the linear equation for y
Substitute into the quadratic equation
Solve for x
Substitute back into the linear equation for y
Write the conclusion

Symmetric systems: call in Vieta's formulas

Some systems are especially neat: swap $x$ and $y$ and the system stays the same. The most basic one is

{x + y = s x y = p

We know the sum and the product of the two numbers. Vieta's formulas say these two numbers are exactly the roots of $t^{2} - s t + p = 0$ . So there is no need to substitute: just solve this quadratic equation.

Predict first

How many real solutions does the system $x + y = 4$ , $x y = 5$ have?

0

1

2

Choose an answer first; then you can see the result.

Notice that the solutions of a symmetric system always come in pairs: if $(a, b)$ is a solution, then so is $(b, a)$ . In the picture, they are reflections of each other in the line $y = x$ .

Both quadratic

Two quadratic curves can have as many as four common points. Now there is no linear equation ready to substitute, so we have to find a way to make one.

The circle

x^{2} + y^{2} = 25

and the hyperbola

x y = 12

: four common points.

One problem, several solutions

Solve the system $x^{2} + y^{2} = 25$ , $x y = 12$ .

Make perfect squares

Add $2$ times the second equation to the first: $x^{2} + 2 x y + y^{2} = 49$ , that is, $(x + y)^{2} = 49$ , so $x + y = \pm 7$ .

Subtract $2$ times the second equation from the first: $(x - y)^{2} = 1$ , so $x - y = \pm 1$ .

Each of the four combinations is a system of two linear equations:

$x + y = 7$ , $x - y = 1$ : $(4, 3)$
$x + y = 7$ , $x - y = - 1$ : $(3, 4)$
$x + y = - 7$ , $x - y = 1$ : $(- 3, - 4)$
$x + y = - 7$ , $x - y = - 1$ : $(- 4, - 3)$

CostYou need to spot the x²+y²±2xy structure, but after that it is only linear equations.

Substitute to get a biquadratic equation

From $x y = 12$ we know $x \neq = 0$ , so $y = \frac{12}{x}$ . Substitute into the first equation:

x^{2} + \frac{144}{x ^{2}} = 25 \Rightarrow x^{4} - 25 x^{2} + 144 = 0

Treat $x^{2}$ as one block: $(x^{2} - 9) (x^{2} - 16) = 0$ , so $x = \pm 3$ or $x = \pm 4$ .

Then $y = \frac{12}{x}$ gives four solutions: $(3, 4)$ , $(- 3, - 4)$ , $(4, 3)$ , $(- 4, - 3)$ .

CostNo need to hunt for structure, but you must solve a quartic equation.

Practice

Exercise 1Basic

Solve the system $x + y = 5$ , $x y = 6$ . Give all possible values of $x$ . (Separate them with commas.)

Exercise 2Basic

Solve the system $y = x + 1$ , $x^{2} + y^{2} = 13$ . Give all possible values of $x$ .

Exercise 3Advanced

The line $y = x + m$ and the circle $x^{2} + y^{2} = 8$ have exactly one common point. Find all possible values of $m$ .

Exercise 4Challenge

Solve the system $x^{2} + y^{2} = 20$ , $x y = 8$ . Give all possible values of $x$ .

What this move does

Work through one

Example: solve 2x − y = 5, x² + y² = 10

Symmetric systems: call in Vieta's formulas

Both quadratic

Make perfect squares

Substitute to get a biquadratic equation

Practice

Mistake Museum · Exhibits from this stop

Finding y by substituting x back into the quadratic equation

Dividing both sides by an expression with an unknown and losing a solution