The hardest SAT math question types, and how to beat them

Example 1: two constants, infinitely many solutions

Problem The system $6x - 4y = 10$ and $ax + 2y = b$ has infinitely many solutions, where a and b are constants. What is $a + b$?

1. Infinitely many solutions means the second equation is a multiple of the first. The y coefficient goes from $-4$ to 2, so the multiplier is $-\frac{1}{2}$.
   $$-\frac{1}{2}(6x - 4y) = -\frac{1}{2}(10)$$
2. Simplify.
   $$-3x + 2y = -5$$
3. Match it to $ax + 2y = b$: $a = -3$ and $b = -5$.

Answer $a + b = -8$

Example 2: discriminant, then the solution

Problem The equation $2x^2 + bx + 18 = 0$, where b is a positive constant, has exactly one real solution. What is that solution?

1. One real solution means the discriminant is 0.
   $$b^2 - 4(2)(18) = 0$$
2. So $b^2 = 144$, and b is positive.
   $$b = 12$$
3. With a zero discriminant, the solution is $-\frac{b}{2a}$.
   $$x = -\frac{12}{2(2)} = -3$$
4. Check: $2x^2 + 12x + 18 = 2(x + 3)^2$, which is 0 only at $x = -3$.

Answer $x = -3$

Example 3: a circle with an unknown constant

Problem The graph of $x^2 + y^2 + 6x - 2y = k$ is a circle with radius 5. What is k?

1. Complete the square: add 9 for the x terms and 1 for the y terms, on both sides.
   $$(x + 3)^2 + (y - 1)^2 = k + 10$$
2. The right side must be $r^2 = 25$.
   $$k + 10 = 25$$
3. Solve.
   $$k = 15$$

Answer $k = 15$

Example 4: a quadratic from its intercepts

Problem The function $f(x) = a(x - 3)(x + 5)$ has a y-intercept of $-30$. What is the minimum value of f?

1. The y-intercept is $f(0)$.
   $$a(0 - 3)(0 + 5) = -30$$
2. So $-15a = -30$.
   $$a = 2$$
3. The vertex is halfway between the zeros 3 and $-5$.
   $$x = \frac{3 + (-5)}{2} = -1$$
4. Evaluate f at $x = -1$. Since $a > 0$, this is a minimum.
   $$2(-1 - 3)(-1 + 5) = -32$$

Answer The minimum value is $-32$.