SAT Quadratic Equations: The 5-Minute Shortcut

Stop wasting time on the Quadratic Formula when the question doesn't even ask for X.

The Problem

You see this on the SAT:

3x² - 18x + 7 = 0

Question: What is the SUM of the solutions?

🚫 The Trap (What 90% of Students Do)

Most students immediately reach for the Quadratic Formula:

x = (-b ± √(b² - 4ac)) / 2a

They spend 2 minutes calculating both roots, adding them together, and then realize there was a faster way. Time wasted: 90-120 seconds.

✅ The Practix Shortcut (3 Seconds)

Use Vieta's Formula for the sum of roots:

Sum of Roots = -b/a

For the equation 3x² - 18x + 7 = 0:

  • b = -18
  • a = 3
  • Sum = -(-18)/3 = 6

That's it. You're done. The answer is 6.

Step-by-Step Guide

1

Identify a, b, and c

From ax² + bx + c = 0, note that a = 3, b = -18, c = 7.

2

Apply Vieta's Formula

Sum of roots = -b/a = -(-18)/3 = 18/3 = 6

3

Done!

You didn't solve for x₁ or x₂. You went straight to the answer.

Bonus: Product of Roots

If the question asks for the PRODUCT of the roots instead:

Product of Roots = c/a

For our example: 7/3

Try It Yourself

Enter a quadratic equation and get the sum/product instantly.

Sum of Roots
6
Product of Roots
2.33

🎓 The Proof: Why $Sum = -b/a$

For students aiming for an 800, knowing the "trick" isn't enough. Here is the rigorous derivation:

1. Start with the two solutions ($x_1$ and $x_2$) from the Quadratic Formula:

$x_1 = \frac{-b + \sqrt{D}}{2a}, \quad x_2 = \frac{-b - \sqrt{D}}{2a}$

2. Add them together and combine the fractions:

$Sum = \frac{-b + \sqrt{D} - b - \sqrt{D}}{2a}$

3. Notice that $\sqrt{D}$ cancels out perfectly:

$Sum = \frac{-2b}{2a} = -\frac{b}{a}$

🎓 The Proof: Why $Product = c/a$

The Product of Roots follows a similar logic, but uses the "Difference of Squares" pattern:

1. Multiply the two solutions together:

$Product = \left(\frac{-b + \sqrt{D}}{2a}\right) \cdot \left(\frac{-b - \sqrt{D}}{2a}\right)$

2. The numerator is a Difference of Squares $(X+Y)(X-Y) = X^2 - Y^2$:

$Product = \frac{(-b)^2 - (\sqrt{D})^2}{(2a)^2} = \frac{b^2 - D}{4a^2}$

3. Substitute the definition of the discriminant ($D = b^2 - 4ac$):

$Product = \frac{b^2 - (b^2 - 4ac)}{4a^2}$

4. Simplify the numerator (the $b^2$ terms cancel):

$Product = \frac{4ac}{4a^2} = \frac{c}{a}$

The beauty of algebra is that it always simplifies to the truth. — Practix Team

Master All 22 Hard SAT Topics

Get access to interactive simulations, weekly challenges, and the full shortcut library.