Nonlinear equations & systems

Most of these are quadratics. You'll solve them by factoring or the quadratic formula, judge how many solutions they have using the discriminant, and find where a parabola meets a line.

Tested on SAT Advanced Math

What College Board tests

Solving quadratic equations, determining the number of real solutions, and solving systems that pair a line with a parabola. The recurring skill is choosing the right tool: factor when it's clean, use the formula when it isn't, and use the discriminant when the question only asks how many solutions.

Tools for a quadratic

Factor
Fastest when the quadratic factors with whole numbers. Set each factor to 0 and solve.
Formula
— works for every quadratic .
Discriminant
The part under the root, . It alone tells you the number of real solutions.

Discriminant sign: positive → two real solutions; zero → exactly one; negative → no real solutions.


Four worked examples in SAT format. Read the approach, try it yourself, then tap Show the full solution.

1 · Solve by factoring

What are all the solutions to ?

Approach Find two numbers that multiply to the constant (6) and add to the middle coefficient (−5). Those numbers build the factors.

Show the full solution

Answer: B

the given equation
find two numbers: product 6, sum −5
build the factors from those numbers
a product is 0 only if a factor is 0
solve each factor

Why the other choices are wrong

A  Right numbers, wrong signs — would come from , which gives .

C  Multiplies to 6 but adds to 7, not −5.

D  Multiplies to −6, not +6.

2 · Solve with the quadratic formula

What are the solutions to ?

Approach This one doesn't factor cleanly, so use the quadratic formula. Identify , , first, then substitute carefully.

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Answer: A

read the coefficients from
the quadratic formula
substitute; focus on the discriminant first
;
the two solutions from

Why the other choices are wrong

B  Sign-flips both solutions.

C  Comes from mis-factoring rather than using the formula.

D  Doesn't satisfy the equation; check : .

3 · How many real solutions?

In the equation , is a constant. What is the smallest integer value of for which the equation has no real solutions?

Approach "No real solutions" means the discriminant is negative. Build the discriminant in terms of , set it less than 0, and find the smallest integer that works.

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Answer: C

the discriminant, with
no real solutions means the discriminant is negative
solve for
the smallest integer greater than 4

Why the other choices are wrong

A   gives a positive discriminant (two solutions).

B   gives a discriminant of exactly 0 — one solution, not none.

D   works, but isn't the smallest such integer.

4 · Where a line meets a parabola

How many points satisfy both and ?

  1. Zero
  2. Exactly one
  3. Exactly two
  4. Infinitely many

Approach At an intersection both equations give the same , so set them equal. That produces a quadratic — the number of real solutions is the number of intersection points.

Show the full solution

Answer: C

set the two expressions for equal
move everything to one side
factor: product , sum
two -values, so two intersection points
y x (-1,1) (2,4)

The graph confirms it: the line crosses the parabola at and — two points.

Why the other choices are wrong

A  Zero would require the quadratic to have a negative discriminant; here it factors.

B  Exactly one would mean the line is tangent (a repeated root); this has two distinct roots.

D  A line and a parabola can share at most two points, never infinitely many.


Quick reference

Factor
Two numbers: product = constant, sum = middle coefficient.
Formula
when factoring fails.
Discriminant
: + two, 0 one, − none.
Line + parabola
Set equal; the quadratic's roots are the intersections.
Shu's Tutoring SAT · Advanced Math · Nonlinear equations & systems