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.
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
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.
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Answer: B
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
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
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 ?
- Zero
- Exactly one
- Exactly two
- 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.
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Answer: C
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.