A quadratic equation is a second-degree polynomial equation in one variable, typically written as $ax^2 + bx + c = 0$, where $a \\neq 0$.
Explanation
A quadratic equation involves the squared term $x^2$ and can have up to two real roots. Key features include the coefficients $a,b,c$, the discriminant that determines root types, the vertex (maximum or minimum), and a parabolic graph.
Standard form
The standard form is:
$$ax^2 + bx + c = 0$$
Quadratic formula (roots)
The solutions (roots) are given by the quadratic formula:
$$x = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}$$
Derivation (completing the square)
Starting from $ax^2 + bx + c = 0$ (with $a\\neq0$), divide by $a$:
$$x^2 + \\frac{b}{a}x + \\frac{c}{a} = 0$$
Move constant term:
$$x^2 + \\frac{b}{a}x = -\\frac{c}{a}$$
Add $(\\frac{b}{2a})^2$ to both sides:
$$x^2 + \\frac{b}{a}x + \\left(\\frac{b}{2a}\\right)^2 = -\\frac{c}{a} + \\left(\\frac{b}{2a}\\right)^2$$
Left side is a perfect square:
$$\\left(x + \\frac{b}{2a}\\right)^2 = \\frac{b^2 – 4ac}{4a^2}$$
Take square root and solve for $x$ to get the quadratic formula:
$$x = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}$$
Discriminant and nature of roots
Define the discriminant $\Delta = b^2 – 4ac$.
- If $\Delta > 0$: two distinct real roots.
- If $\Delta = 0$: one real repeated root (double root).
- If $\Delta < 0$: two complex conjugate roots.
Methods to solve
- Factoring (when factors are integers or simple).
- Completing the square (derivation method; useful for vertex form).
- Quadratic formula (works always).
- Graphing (visual; finds intercepts of parabola $y=ax^2+bx+c$).
Vertex form and graph
Vertex form: $$y = a(x – h)^2 + k$$ where vertex is $(h,k)$ and axis of symmetry is $x=h$. For $ax^2 + bx + c$, the vertex is at:
$$h = -\\frac{b}{2a},\\quad k = c – \\frac{b^2}{4a}$$
The graph is a parabola opening up if $a>0$ (minimum) and down if $a<0$ (maximum).
Worked examples
1) Factorable example:
Solve $x^2 – 5x + 6 = 0$.
Steps:
- Factor: $(x-2)(x-3)=0$.
- Roots: $x=2$ or $x=3$.
Therefore, the solutions are $x=2$ and $x=3$.
2) Using the quadratic formula:
Solve $x^2 + 4x + 1 = 0$.
Steps:
- Here $a=1,b=4,c=1$. Compute discriminant: $\Delta = 4^2 – 4(1)(1)=12$.
- Apply formula: $$x = \\frac{-4 \\pm \\sqrt{12}}{2} = \\frac{-4 \\pm 2\\sqrt{3}}{2} = -2 \\pm \\sqrt{3}.$$
Therefore, $x = -2 \\pm \\sqrt{3}$.
3) Completing the square:
Solve $2x^2 – 8x + 5 = 0$.
Steps:
- Divide by 2: $x^2 – 4x + \\tfrac{5}{2}=0$.
- Move constant: $x^2 – 4x = -\\tfrac{5}{2}$.
- Add $(\\tfrac{4}{2})^2=4$: $x^2 -4x +4 = -\\tfrac{5}{2} +4 = \\tfrac{3}{2}$.
- So $(x-2)^2 = \\tfrac{3}{2}$, $x-2 = \\pm \\sqrt{\\tfrac{3}{2}}$.
- Solutions: $x = 2 \\pm \\sqrt{\\tfrac{3}{2}}$ (can rationalize if desired).
Applications & tips
Quadratics model projectile motion, area problems, optimization, and many physics/economics situations. If factoring fails quickly, use the quadratic formula. Check the discriminant first to anticipate the type of roots.
Practice problems
- Solve $x^2 + 2x – 8 = 0$.
- Find vertex and axis of symmetry for $y = -3x^2 + 6x -1$.
- Solve $3x^2 + x + 5 = 0$ and classify the roots.
If you want, I can solve the practice problems step-by-step or make a short worksheet with solutions.