ALGEBRA · CORE CONCEPT

Quadratic equations

A quadratic equation contains a squared unknown. Its algebraic roots are exactly the x-intercepts of a parabola, linking formulas, factorisation and geometry.

for \(a\neq0\)

\(ax^2+bx+c=0\)

↓

\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)

1. Standard form

A quadratic equation in one variable can be written as

\[ax^2+bx+c=0,\qquad a\neq0\]

The coefficient \(a\) must be non-zero; otherwise the equation is linear. Dividing by \(a\) gives the monic form \(x^2+px+q=0\), where \(p=b/a\) and \(q=c/a\).

2. Deriving the quadratic formula

The formula follows from completing the square, not from memorisation alone.

\[ax^2+bx+c=0\]

Divide by \(a\):

\[x^2+\frac ba x=-\frac ca\]

Add \(\left(\frac{b}{2a}\right)^2\) to both sides:

\[\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\]

Taking square roots and isolating \(x\) yields

\[\boxed{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}\]

3. The discriminant tells you how many real roots exist

Define \(\Delta=b^2-4ac\). Its sign determines the real-root structure.

\(\Delta>0\)Two distinct real roots.

\(\Delta=0\)One repeated real root, \(x=-b/(2a)\).

\(\Delta<0\)No real roots; over \(\mathbb C\), there are two complex-conjugate roots.

4. Graphical interpretation

The function \(y=ax^2+bx+c\) is a parabola. Solving the quadratic equation asks where this parabola intersects the x-axis.

The vertex has x-coordinate

\[x_v=-\frac{b}{2a}\]

For \(x^2-4x+3=0\), the roots are \(1\) and \(3\), while the vertex is at \((2,-1)\).

5. Which solving method should you use?

Factorisation

Fastest when the polynomial factors cleanly, e.g. \(x^2-5x+6=(x-2)(x-3)\).

Quadratic formula

Universal for every quadratic equation with \(a\neq0\).

Completing the square

Best for deriving the formula, understanding vertex form, or revealing geometry.

Square-root method

Efficient for equations such as \((x-h)^2=k\).

6. Vieta's formulas

If \(x_1,x_2\) are the roots of \(ax^2+bx+c=0\), then

\[x_1+x_2=-\frac ba,\qquad x_1x_2=\frac ca\]

These identities are useful for checking roots, factoring and reasoning about parameters without solving explicitly.

7. Worked examples

Factorisation

\[x^2-5x+6=0\iff(x-2)(x-3)=0\]

Therefore \(x=2\) or \(x=3\).

Quadratic formula

For \(2x^2+3x-2=0\), \(a=2,b=3,c=-2\), so \(\Delta=25\).

\[x=\frac{-3\pm5}{4}\in\left\{\frac12,-2\right\}\]

No real roots

For \(x^2+2x+5=0\), \(\Delta=4-20=-16\). Therefore there are no real roots; over the complex numbers the roots are \(-1\pm2i\).

8. Interactive quadratic solver

Enter coefficients for \(ax^2+bx+c=0\).

abc

x₁ = 1, x₂ = 3

Discriminant Δ = 4

9. Practice

\(x^2+8x+15=0\)
\(3x^2-6x+3=0\)
\(x^2+2x+5=0\)
\(2x^2-7x+3=0\)

Show answers

1) \(-3,-5\). 2) \(1\) repeated. 3) No real roots; complex roots \(-1\pm2i\). 4) \(3,\frac12\).