The Quadratic Formula
Any equation of the form Ax² + Bx + C = 0 can be solved using the quadratic formula. The key insight is that the discriminant—the expression under the square root—tells you exactly how many and what type of solutions exist.
x = (−B ± √Δ) / 2A
where Δ = B² − 4AC
A— Coefficient of x² (cannot be zero, or the equation is linear, not quadratic)B— Coefficient of xC— Constant termΔ (discriminant)— Determines the nature and number of solutions: positive means two real roots, zero means one repeated root, negative means two complex conjugate roots
Understanding the Discriminant
The discriminant is the expression B² − 4AC that appears under the square root in the quadratic formula. Its value tells you everything about the solutions before you even calculate them:
- Δ > 0: The equation has two distinct real solutions. For example,
x² − 5x + 6 = 0has discriminant 25 − 24 = 1, yielding rootsx = 2andx = 3. - Δ = 0: The equation has exactly one real solution (a repeated root). This parabola just touches the x-axis. For instance,
x² − 2x + 1 = 0has discriminant 4 − 4 = 0 and one solutionx = 1. - Δ < 0: The equation has no real solutions; instead, there are two complex conjugate roots. The parabola never crosses the x-axis. For example,
x² + x + 1 = 0has discriminant 1 − 4 = −3, giving complex solutions involving the imaginary uniti.
How to Use the Formula Step-by-Step
To solve 3x² − 7x + 2 = 0 using the quadratic formula, follow these steps:
- Identify your coefficients:
A = 3,B = −7,C = 2. - Calculate the discriminant:
Δ = (−7)² − 4(3)(2) = 49 − 24 = 25. - Check the sign: Since Δ = 25 > 0, you have two real solutions.
- Apply the formula:
x = (7 ± √25) / 6 = (7 ± 5) / 6, givingx₁ = 2andx₂ = 1/3.
Always first rearrange your equation into standard form Ax² + Bx + C = 0 before identifying coefficients. Move all terms to one side and combine like terms.
Complex Solutions and the Imaginary Unit
When the discriminant is negative, you cannot take a real square root, but you can work with complex numbers. Complex solutions are written as p + qi, where p is the real part, q is the imaginary part, and i = √(−1).
For an equation with a negative discriminant, the two solutions are always complex conjugates: if one solution is p + qi, the other is p − qi. For example, solving x² + 2x + 5 = 0 yields Δ = 4 − 20 = −16. The real part is −2/2 = −1, and the imaginary part is ±√16/2 = ±2, giving solutions −1 + 2i and −1 − 2i.
Common Pitfalls and Practical Tips
Avoid these frequent mistakes when applying the quadratic formula.
- Don't forget to rearrange first — Your equation must be in the form <code>Ax² + Bx + C = 0</code> with zero on the right side. If you see <code>3x² + x = 5</code>, rewrite it as <code>3x² + x − 5 = 0</code> before identifying coefficients. Errors at this stage propagate through the entire calculation.
- Watch the signs of your coefficients — Include the sign as part of each coefficient. In <code>2x² − 3x − 4 = 0</code>, you have <code>A = 2</code>, <code>B = −3</code>, and <code>C = −4</code>, not <code>B = 3</code> and <code>C = 4</code>. Incorrect signs in the formula lead to completely wrong roots.
- Check whether A equals zero — If <code>A = 0</code>, you don't have a quadratic equation—you have a linear equation. The quadratic formula will fail because you'll divide by zero. For example, <code>0x² + 5x + 3 = 0</code> simplifies to <code>5x + 3 = 0</code>, solved as <code>x = −3/5</code> using basic algebra.
- Handle complex solutions carefully — When Δ < 0, ensure you correctly compute the imaginary part as <code>±√|Δ| / 2A</code> (using the absolute value of Δ). Many calculators handle this automatically, but if calculating by hand, remember that <code>√(−16) = 4i</code>, not <code>−4i</code>.