Linear and quadratic systems

How the SAT tests this, with worked examples — plus the 4 skills it covers. Free to read, no account needed.

Here one equation is a line and the other a parabola, and the solutions are the points where they meet. A line can cut a parabola twice, touch it once, or miss it entirely.

The method is substitution every time. Put the linear expression for

y

into the quadratic, collect everything on one side, and you have an ordinary quadratic equation to solve.

How many solutions there are is decided by the discriminant

b^2 - 4ac

of that resulting quadratic. Positive gives two intersections, zero gives exactly one — the tangent case — and negative gives none.

That is why questions asking for the value of a constant making the system have exactly one solution are really asking you to set the discriminant to zero and solve.

Remember to finish the job. Solving gives you

x

; the question usually wants the point, so substitute back into the linear equation to get

y

Worked examples

Solve

y = x^2

and

y = 0.5x + 3

.
Substitute and collect

{x^2 - 0.5x - 3 = 0}

which factors to give

x = 2

and

x = -1.5

. Substituting back gives the points

(2, 4)

and

(-1.5, 2.25)

For what value of

c

does

y = x^2

and

y = 2x + c

meet exactly once?
Substituting gives

x^2 - 2x - c = 0

, whose discriminant is

4 + 4c

. One solution needs the discriminant to be zero

{4 + 4c = 0 \;\Rightarrow\; c = -1}

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Linear and quadratic systems

Worked examples

Is linear and quadratic systems costing you points?

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