A quadratic equation can always be solved by completing the square or by the quadratic formula, but the shortest method depends on the equation’s visible structure. Factoring is efficient when a product can be exposed; the square-root method fits an isolated square; completing the square reveals the vertex; and the quadratic formula handles arbitrary coefficients. This editable Algebra I and Algebra II lesson develops those connections instead of presenting four unrelated procedures.
The central connection
For $a\ne0$, completing the square gives
$$ ax^2+bx+c =a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}. $$
Setting the expression equal to zero and isolating the square produces
$$ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. $$
Thus the quadratic formula is the general completed-square argument. The same identity places the vertex at
$$ \left(-\frac{b}{2a},-\frac{b^2-4ac}{4a}\right), $$
which explains graphically why the discriminant $\Delta=b^2-4ac$ controls the roots. For an upward-opening parabola, $\Delta>0$ places the vertex below the axis and produces two intercepts; $\Delta=0$ places it on the axis; and $\Delta<0$ places it above the axis.
Which method fits?
| Visible structure | Efficient method | Mathematical reason |
|---|---|---|
| $(mx+n)(px+q)=0$ | Factoring | A zero product forces at least one factor to be zero. |
| $(x-h)^2=k$ | Square roots | Both inverse images, $\pm\sqrt{k}$, are immediately available. |
| $x^2+bx=c$ | Completing the square | Adding $(b/2)^2$ creates a perfect square. |
| $ax^2+bx+c=0$ with no evident factors | Quadratic formula | The formula is systematic for every $a\ne0$. |
The equation $x^2-6x+5=0$, for example, can be written as either
$$ (x-1)(x-5)=0 $$
or
$$ (x-3)^2=4. $$
Both forms give roots $1$ and $5$. The factored form is shorter for solving; the completed-square form also gives the vertex $(3,-4)$.
Representative problems
- Solve $6x^2-x-2=0$ exactly and verify the sum and product of the roots.
- Complete the square in $3x^2-12x+7$ and state the axis and minimum value.
- Find all real $k$ for which $x^2+kx+9=0$ has exactly one real solution.
- Explain why $x(x-4)=12$ does not permit the immediate conclusion $x=0$ or $x=4$.
- A rectangle beside a wall uses $80$ feet of fencing on three sides and has area $600$ square feet. Determine every possible pair of dimensions.
- Prove that, for $a>0$ and distinct real roots $r_1<r_2$, the inequality $f(k)<0$ holds exactly when $r_1<k<r_2$.
Included resource
The editable project contains a textbook-style student lesson, forty-two mixed problems, a separately distributable teacher key, a geometric completing-square diagram, a three-case discriminant graph, and a visual map of standard, factored, and vertex forms. The source uses US Letter paper and a tagged-PDF output target; school identity, exercise selection, page furniture, and examples remain editable.
Curriculum alignment and limits
The lesson addresses CCSS HSA-REI.B.4, including completing the square, deriving and using the quadratic formula, and recognizing complex solutions. It also supports HSA-SSE.B.3 through purposeful form changes and HSF-IF.C.7a through graph interpretation.
The package remains a draft pending independent mathematics and classroom review. It does not cover conic-section geometry, general polynomial root theory, or advanced conditions for locating roots in prescribed intervals. Projectile examples use an ideal constant-gravity model and omit air resistance.