Math FoundationsEquationsQuadratic Equations

Quadratic formula

A clear derivation of the quadratic formula with a detailed explanation of every step. Learn what the discriminant is, where it comes from, and how it tells you how many roots a quadratic equation has.
Key elements:
Connections:
The quadratic formula is derived by completing the square, so you really need to know how that works!
Adding and compensating
Statistics:
Term2
Statement1
Important2
Problem14
Article
Summary
Practice

General root formulas

General formulas for solutions (roots) let you basically not solve the equation step by step. You just substitute the coefficients and get the roots directly! This is especially useful for harder equations like quadratics, where the usual method takes a whole chain of steps.
How to solve any quadratic equation
A direct derivation of the quadratic formula via completing the square. If you want a more detailed explanation, the article walks through every step.
ActionEquation
1Write it in the general form.Ax2+Bx+C=0\displaystyle Ax^2 + Bx + C = 0
2Divide both sides by A.x2+BAx+CA=0\displaystyle x^2 + \frac{B}{A}x + \frac{C}{A} = 0
3Move the constant term to the right.x2+BAx=−CA\displaystyle x^2 + \frac{B}{A}x = -\frac{C}{A}
4Add and compensate the 2 needed for completing the square.x2+2⋅xBA⋅12=−CA\displaystyle x^2 + \yellow{2} \cdot x \frac{B}{A} \cdot \yellow{\frac{1}{2}} = -\frac{C}{A}
5Add and compensate b2b^2 to complete the square.x2+2⋅x⋅B2A+(B2A)2⏟a2+2ab+b2=(a+b)2−(B2A)2=−CA\displaystyle \underbrace{x^2 + 2\cdot x \cdot \frac{B}{2A} + \yellow{\left( \frac{B}{2A} \right)^2}}_{\small a^2 + 2ab + b^2 = (a+b)^2} - \yellow{\left( \frac{B}{2A} \right)^2} = -\frac{C}{A}
6Complete the square on the left.(x+B2A)2=−CA+(B2A)2\displaystyle \left( x + \frac{B}{2A} \right)^2 = -\frac{C}{A} + \left( \frac{B}{2A} \right)^2
7Put the right-hand side over a common denominator.(x+B2A)2=B2−4AC4A2\displaystyle \left( x + \frac{B}{2A} \right)^2 = \frac{B^2 - 4AC}{4A^2}
8Introduce the discriminant.(x+B2A)2=D4A2\displaystyle \left( x + \frac{B}{2A} \right)^2 = \frac{\brand{D}}{4A^2}
9Take the square root of both sides (if D≥0D \ge 0).x+B2A=±D2A\displaystyle x + \frac{B}{2A} = \pm \frac{\sqrt{D}}{2A}
10Isolate x and get the final formula.x=−B±D2A\displaystyle x = \frac{-B \pm \sqrt{D}}{2A}

Examples: solving quadratic equations

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Example
Solve the quadratic equation using the quadratic formula.
x2−2x−8=0x^2 - 2x - 8 = 0

Biquadratic equation

A special type of fourth-degree equation that can be written in the general form:
Ax4+Bx2+C=0,A≠0Ax^4 + Bx^2 + C = 0, \quad A \neq 0
Examples:
x4+16x2+55=0x^4 + 16x^2 + 55 = 0
12y2−3y4=1 \frac{1}{2}y^2 - \sqrt{3}y^4 = 1
5=z(z3−23z) 5 = z(z^3 - 2\sqrt{3}z)
Such equations can have up to four roots, and they are very easy to solve by reducing them to a quadratic via the substitution t=x2t = x^2:
At2+Bt+C=0At^2 + Bt + C = 0

Examples: solving biquadratic equations

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Elementary
Solve the biquadratic equation:
2x4−3x2+1=02x^4 - 3x^2 + 1 = 0
Quadratic formula