SAT Math: Nonlinear equations and systems

Nonlinear equations in one variable and systems in two variables.

Advanced Math · Nonlinear equations in one variable and systems of equations in two variables
Both modules
20 practice questions · 5 with free solutions
Calculator allowed (Desmos)

Practise nonlinear equations and systems →Adaptive SAT module

What the test covers

Quadratics (factor, formula, discriminant), radical and rational equations with extraneous roots
Line-parabola systems

Key ideas

Quadratics: factorise, complete the square or use $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ . The sum of the roots is $- \frac{b}{a}$ .
The discriminant $b^{2} - 4 a c$ decides whether there are 2, 1 or 0 real solutions.
Radical and rational equations can produce extraneous solutions: always check in the original equation.
A line and a parabola meet at 0, 1 or 2 points: substitute and use the discriminant.

Common mistakes

Squaring $\sqrt{x + 7} = x - 5$ creates a false root; check both.
A denominator can never be zero.

Do it in Desmos

Graph both sides and count or read the intersections.

IB link: IB SL 2.7.

Worked example

What is the sum of the solutions to $2 x^{2} - 7 x + 3 = 0$ ?

$\frac{3}{2}$
$\frac{7}{2}$
$7$
$- \frac{7}{2}$

Answer: B: $\frac{7}{2}$

$(2 x - 1) (x - 3) = 0$ , so $x = \frac{1}{2}$ or $3$ ; the sum is $\frac{7}{2}$ .

Practice questions

Try each question before opening the solution. Desmos or your own calculator is allowed on every SAT Math question. Solutions to the first 4 are open to everyone; the rest need only a free account. Every answer here was checked independently before it was published.

Question 1

What is the sum of the solutions to $2 x^{2} + 3 x - 2 = 0$ ?

$- 1$
$2$
$- \frac{3}{2}$
$\frac{3}{2}$

Show the answer and solution

Answer: C: $- \frac{3}{2}$

The solutions are $- 2$ and $\frac{1}{2}$ (factor, or use sum $= - \frac{b}{a}$ ). Their sum is $- \frac{3}{2}$ .

Question 2

In the equation $x^{2} - 6 x + c = 0$ , $c$ is a constant. The equation has exactly one real solution. What is the value of $c$ ?

Student-produced response: type your answer.

Show the answer and solution

Answer: 9

One real solution means the discriminant is zero: $6^{2} - 4 c = 0$ , so $c = 9$ .

Question 3

What are all the solutions to $\sqrt{x + 7} = x - 5$ ?

There are no solutions
$2$ and $9$
$9$ only
$2$ only

Show the answer and solution

Answer: C: $9$ only

Squaring: $x + 7 = x^{2} - 10 x + 25$ , $x^{2} - 11 x + 18 = 0$ , $(x - 2) (x - 9) = 0$ . Check: $x = 9$ : $\sqrt{16} = 4 = 9 - 5$ ✓. $x = 2$ : $\sqrt{9} = 3 \neq - 3$ ✗ (created by squaring). Only 9.

Question 4

The line $y = k x - 1$ , where $k$ is a positive constant, intersects the parabola $y = x^{2} + 15$ at exactly one point. What is the value of $k$ ?

Student-produced response: type your answer.

Show the answer and solution

Answer: 8

$x^{2} + 15 = k x - 1$ gives $x^{2} - k x + 16 = 0$ . Exactly one intersection: $k^{2} - 64 = 0$ , $k = \pm 8$ . As $k > 0$ , $k = 8$ .