--- title: Solving Quadratic Equations subtitle: Choosing and connecting the methods author: Mathematics Department organization: School or district date: 2026-09-11 description: A connected treatment of factoring, square roots, completing the square, the quadratic formula, the discriminant, complex roots, and quadratic models. keywords: [quadratic equations, factoring, completing the square, quadratic formula, discriminant, complex roots] template: assignment paper: us-letter titlePage: false toc: false headingNumbers: false equationNumbers: false columns: 1 figurePlacement: none tableCaptionPosition: top header: Quadratic Equations | Connected Methods footer: Textbook lesson, practice, and reasoning pdfProfile: tagged language: en-US --- # Quadratic equations and their roots A **quadratic expression** in $x$ has the form $ax^2+bx+c$, where $a$, $b$, and $c$ are real numbers and $a\ne0$. The condition $a\ne0$ is essential: otherwise the term of degree two disappears. An expression has a value but makes no assertion. The formula $f(x)=ax^2+bx+c$ defines a **quadratic function**, while :::equation{#eq:general alt="a x squared plus b x plus c equals zero, where a is not zero" numbered=true} \[ ax^2+bx+c=0\text{ with }a\ne0, \] ::: is a **quadratic equation**. A number $r$ is a solution, or **root**, of @eq:general when substitution makes the equation true. The same number is a **zero** of $f$ because $f(r)=0$, and the point $(r,0)$ is an $x$-intercept of the graph $y=f(x)$. These are three descriptions of one fact. Solving an equation means finding its entire solution set. Each algebraic change must therefore produce an equivalent equation. Adding the same expression to both sides, multiplying both sides by a nonzero number, and replacing an expression by an equal expression preserve the solution set. Squaring both sides need not preserve it and can introduce extraneous solutions; taking a square root requires both signs. ## One function, three useful forms A quadratic function often appears in three forms. :::table{#tbl:forms caption="Equivalent forms reveal different features of the same quadratic function"} | Form | Formula | Information visible immediately | | --- | --- | --- | | Standard | $a x^2+b x+c$ | coefficients, $y$-intercept $c$, discriminant $b^2-4ac$ | | Factored | $a(x-r_1)(x-r_2)$ | roots $r_1,r_2$ and $x$-intercepts | | Vertex | $a(x-h)^2+k$ | vertex $(h,k)$, axis $x=h$, minimum or maximum $k$ | ::: :::figure{#fig:forms caption="The form of a quadratic changes what can be read without further algebra" alt="A triangular map connecting standard, factored, and vertex forms; standard reveals coefficients, factored reveals roots, and vertex form reveals the turning point" width=84 placement=none} ![Standard, factored, and vertex forms connected in a triangular map](quadratic-forms-map.svg) ::: For example, $$x^2-2x-8=(x-4)(x+2)=(x-1)^2-9.$$ The standard form gives the $y$-intercept $-8$. The factored form gives roots $4$ and $-2$. The vertex form gives the vertex $(1,-9)$ and the axis of symmetry $x=1$. Expanding either transformed form recovers the standard form, so all three describe the same function. # Factoring and the zero-product principle Factoring replaces a sum by a product. This is useful because the real numbers have the **zero-product property**: > [!IMPORTANT] > If $uv=0$, then $u=0$ or $v=0$. Conversely, if either factor is zero, then their product is zero. The forward implication follows because if $u\ne0$, division by $u$ gives $v=0$; if $u=0$, the first alternative already holds. No corresponding rule exists for a sum: $u+v=0$ does not force either term to be zero. **Example 1.** The equation $6x^2-x-2=0$ factors as $$ (3x-2)(2x+1)=0.$$ The zero-product property gives $3x-2=0$ or $2x+1=0$, hence $$x=\frac23\quad\text{or}\quad x=-\frac12.$$ Both values satisfy the original equation. The factorization is especially efficient because the integer product structure is visible. **Example 2.** The equation $2x^2+10x=0$ contains a common factor: $$2x(x+5)=0.$$ Its roots are $0$ and $-5$. Dividing the original equation by $x$ would have discarded the root $x=0$; factoring preserves it. Factoring is exact and short when the factors are recognizable. It is not a universal search procedure. A quadratic such as $x^2-8x+3$ has irrational roots and does not factor into linear factors with integer coefficients. # Isolating a square If an equation has the form $(x-h)^2=k$, its structure already contains a square. > [!IMPORTANT] > For $k\ge0$, $z^2=k$ if and only if $z=\sqrt{k}$ or $z=-\sqrt{k}$. Thus $z=\pm\sqrt{k}$. For $k<0$, there is no real solution. Both signs are necessary because opposite numbers have the same square. The symbol $\sqrt{k}$ denotes the nonnegative square root; the symbol $\pm$ supplies the two possible values of the quantity being squared. **Example 3.** From $3(x-2)^2=27$, $$ (x-2)^2=9\text{ implies }x-2=\pm3\text{ and }x=5\text{ or }x=-1.$$ **Example 4.** The equation $(2x+1)^2=7$ gives $$2x+1=\pm\sqrt7\text{ and }x=\frac{-1\pm\sqrt7}{2}.$$ If complex numbers are admitted, a negative right side also has solutions. Since $i^2=-1$, the equation $z^2=-k$ with $k>0$ gives $z=\pm i\sqrt{k}$. # Completing the square Completing the square changes standard form into vertex form. Its source is the identity $$ (x+p)^2=x^2+2px+p^2.$$ Therefore $x^2+qx$ becomes a perfect square when $(q/2)^2$ is added: :::equation{#eq:monic-completion alt="x squared plus q x equals the quantity x plus q over two squared minus q squared over four" numbered=true} \[ x^2+qx=\left(x+\frac q2\right)^2-\frac{q^2}{4}. \] ::: The geometry is visible in @fig:area. A square of side $x$ and two rectangles, each with one side $p$, leave a corner of area $p^2$. Adding that corner creates a square of side $x+p$. :::figure{#fig:area caption="The area identity x squared plus 2 p x plus p squared equals the quantity x plus p squared" alt="A large square of side x plus p divided into an x by x square, two x by p rectangles, and a p by p corner square" width=58 placement=none} ![Area model for completing the square](completing-square-area.svg) ::: For a general quadratic, factor $a$ from the terms that contain $x$ and apply @eq:monic-completion with $q=b/a$: :::equation{#eq:completion alt="a x squared plus b x plus c equals a times the quantity x plus b over two a squared minus the discriminant over four a" numbered=true} \[ \begin{aligned} ax^2+bx+c &=a\left(x^2+\frac ba x\right)+c\\ &=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\\ &=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2-4ac}{4a}. \end{aligned} \] ::: Writing $\Delta=b^2-4ac$, the vertex is $$\left(-\frac{b}{2a},-\frac{\Delta}{4a}\right).$$ If $a>0$, the squared term in @eq:completion is nonnegative and the vertex value is the minimum. If $a<0$, it is the maximum. The formula also gives the axis of symmetry $x=-b/(2a)$. **Example 5.** Completing the square in $x^2+6x-7=0$ gives $$ (x+3)^2-16=0\text{ and }(x+3)^2=16,$$ so $x=1$ or $x=-7$. **Example 6.** For $2x^2-4x-1=0$, $$2(x^2-2x)-1=0\text{ and }2\bigl((x-1)^2-1\bigr)-1=0.$$ Hence $2(x-1)^2=3$, and $$x=1\pm\sqrt{\frac32}=1\pm\frac{\sqrt6}{2}.$$ The nonmonic case requires care: the compensating square is added inside a quantity multiplied by $a$. # The quadratic formula Completing the square solves every quadratic equation and produces a single general formula. Begin with @eq:general and divide by $a$, which is legal because $a\ne0$: $$x^2+\frac ba x=-\frac ca.$$ Adding $\left(\frac{b}{2a}\right)^2$ to both sides gives $$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}.$$ Taking both square roots and isolating $x$ yields :::equation{#eq:formula alt="x equals negative b plus or minus the square root of b squared minus four a c, all over two a" numbered=true} \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \] ::: The quadratic formula is not a separate fact unrelated to the other methods. It is the completed-square argument applied once to the general equation. **Example 7.** For $2x^2-4x-1=0$, $a=2$, $b=-4$, and $c=-1$. Then :::equation{#eq:formula-example alt="x equals four plus or minus the square root of twenty-four over four, which equals one plus or minus the square root of six over two" numbered=false} \[ \begin{aligned} x&=\frac{4\pm\sqrt{(-4)^2-4(2)(-1)}}{4}\\ &=\frac{4\pm\sqrt{24}}{4}\\ &=1\pm\frac{\sqrt6}{2}. \end{aligned} \] ::: which agrees with Example 6. **Example 8.** For $5x^2+2x-3=0$, $$x=\frac{-2\pm\sqrt{4+60}}{10}=\frac{-2\pm8}{10},$$ so $x=3/5$ or $x=-1$. Although the formula works, factoring as $(5x-3)(x+1)=0$ is shorter. # The discriminant and the graph The quantity $$\Delta=b^2-4ac$$ is the **discriminant**. It appears under the square root in @eq:formula and controls the number of real roots. :::table{#tbl:discriminant caption="The discriminant classifies the roots of a real quadratic equation"} | Condition | Real solutions | Graph of $y=ax^2+bx+c$ | | --- | --- | --- | | $\Delta>0$ | two distinct real roots | crosses the $x$-axis twice | | $\Delta=0$ | one repeated real root $-b/(2a)$ | touches the $x$-axis at the vertex | | $\Delta<0$ | no real roots; two complex conjugates | does not meet the $x$-axis | ::: :::figure{#fig:discriminant caption="Three upward-opening parabolas illustrate positive, zero, and negative discriminants" alt="Three coordinate plots: one parabola crosses the x-axis twice, one touches it once at its vertex, and one remains above it" width=94 placement=none} ![Parabolas with two, one, and no real x-intercepts](discriminant-parabolas.svg) ::: This classification is also encoded in the vertex. For $a>0$, the vertex height is $-\Delta/(4a)$. It lies below, on, or above the $x$-axis according as $\Delta$ is positive, zero, or negative. For $a<0$ the orientation reverses, but the same three intersection counts remain. **Example 9.** The equation $4x^2+12x+9=0$ has $\Delta=144-144=0$, so it has one repeated root. Indeed, $$4x^2+12x+9=(2x+3)^2,$$ and $x=-3/2$. **Example 10.** For $x^2-4x+13=0$, $\Delta=16-52=-36$. There are no real roots. Since $\sqrt{-36}=6i$, $$x=\frac{4\pm6i}{2}=2\pm3i.$$ Because the coefficients are real, nonreal roots occur as conjugates: changing $i$ to $-i$ leaves their sum and product real. # Choosing an efficient method Every method is valid when its algebra is valid, but not every method is equally revealing or economical. :::table{#tbl:choice caption="Structural clues for selecting a solving method"} | Visible structure | Usually efficient | Reason | | --- | --- | --- | | Product equal to zero | zero-product property | each factor can be set equal to zero | | Isolated square $(x-h)^2=k$ | square-root method | the equation is already in inverse-operation form | | Easily factorable trinomial | factoring | roots appear without radicals or extended algebra | | Vertex or extremum is also needed | completing the square | the solving work simultaneously exposes the vertex | | Arbitrary coefficients or uncertain factorization | quadratic formula | it is systematic and always applies | | Approximate roots only | graph or numerical method | useful for estimation, not a substitute for exact roots | ::: Method choice is a claim about efficiency, not about correctness. The equation $x^2-6x+5=0$ can be factored as $(x-1)(x-5)=0$, completed to $(x-3)^2=4$, or placed in the quadratic formula. Each gives $x=1$ or $x=5$; the factorization is shortest, while completion also reveals the vertex $(3,-4)$. > [!NOTE] > A graph can verify the number and approximate location of real roots, but a screen image normally cannot establish an exact irrational value. Exact algebra and graphical evidence answer different questions. ## Verification Substitution is decisive. If $r$ is proposed as a root of $ax^2+bx+c=0$, evaluating $ar^2+br+c$ must give zero. For factored form, the product should expand to the original expression. For a graph, its intercept count must agree with the sign of the discriminant. Vieta's relations provide a second check. If $r_1$ and $r_2$ are the roots, then $$a(x-r_1)(x-r_2)=ax^2-a(r_1+r_2)x+ar_1r_2.$$ Comparison with $ax^2+bx+c$ gives :::equation{#eq:vieta alt="r one plus r two equals negative b over a, and r one times r two equals c over a" numbered=true} \[ r_1+r_2=-\frac ba\text{ and }r_1r_2=\frac ca. \] ::: # Quadratic models A model begins with a quantity and its constraints, not merely an equation. Algebraic roots that violate those constraints are rejected in context. **Example 11: area.** A rectangle has area $54$ square units and its length is $3$ units greater than its width. If the width is $w>0$, then $$w(w+3)=54\text{ and }w^2+3w-54=0.$$ Factoring gives $(w+9)(w-6)=0$. The algebraic roots are $-9$ and $6$, but a width cannot be negative. The rectangle is $6$ units by $9$ units. **Example 12: vertical motion.** In feet, a ball launched from height $5$ with initial vertical velocity $48$ feet per second has idealized height $$h(t)=-16t^2+48t+5.$$ Its maximum occurs at $t=-48/(2\cdot-16)=3/2$ seconds and is $h(3/2)=41$ feet. It reaches the ground when $h(t)=0$: :::equation{#eq:flight-time alt="t equals negative forty-eight plus or minus the square root of forty-eight squared minus four times negative sixteen times five, divided by negative thirty-two, which equals six plus or minus the square root of forty-one over four" numbered=false} \[ \begin{aligned} t&=\frac{-48\pm\sqrt{48^2-4(-16)(5)}}{-32}\\ &=\frac{6\pm\sqrt{41}}{4}. \end{aligned} \] ::: The negative root represents a time before launch and is outside the model. The physical time is $(6+\sqrt{41})/4\approx3.10$ seconds. # Errors that change the solution set - From $z^2=25$, writing $z=5$ omits the solution $z=-5$. - From $uv=0$, dividing by $u$ before considering $u=0$ can discard a root. - Factoring when the equation is not equal to zero does not justify setting factors equal to zero. - In $a(x-h)^2+k$, the vertex is $(h,k)$; the sign inside the parentheses is opposite the vertex coordinate. - The discriminant is $b^2-4ac$. Parentheses are essential when $b$ or $c$ is negative. - The entire numerator $-b\pm\sqrt{\Delta}$ is divided by $2a$ in the quadratic formula. - A decimal from a graph is evidence for location, not an exact form of an irrational root. - A mathematically valid negative root may be invalid for a length, elapsed time, or count. # Exercises Exact values are required unless an approximation is requested. Reasoning is part of each answer. ## Structure and method 1. State whether each object is an expression, a function definition, or an equation: $3x^2-5x+1$; $g(x)=3x^2-5x+1$; $3x^2-5x+1=0$. 2. For $f(x)=2(x-3)(x+1)$, give the roots, the $x$-intercepts, and the $y$-intercept. 3. For $g(x)=-3(x+2)^2+12$, give the vertex, axis of symmetry, maximum value, and roots. 4. Write $x^2-2x-8$ in standard, factored, and vertex forms. State one feature visible in each. 5. Select an efficient exact method for each equation and justify the selection without solving: $(4x-1)(x+7)=0$; $5(x-2)^2=45$; $x^2+10x+7=0$; $7x^2-3x-11=0$. 6. Explain why $x(x-4)=12$ does not permit the immediate conclusion $x=0$ or $x=4$. Rewrite it in a form to which the zero-product property applies. ## Exact solutions ::::regions{tracks="1 1" gutter=18} :::region 7. $x^2-9=0$ 8. $3(x-2)^2=27$ 9. $x^2+7x+12=0$ 10. $6x^2-x-2=0$ 11. $2x^2+10x=0$ 12. $x^2+6x-7=0$ ::: :::region 13. $x^2-8x+3=0$ 14. $2x^2-4x-1=0$ 15. $5x^2+2x-3=0$ 16. $4x^2+12x+9=0$ 17. $x^2-4x+13=0$ 18. $3x^2+6x+7=0$ ::: :::: ## Forms, vertices, and roots 19. Complete the square in $x^2-10x+7$ and state its minimum value. 20. Write $3x^2-12x+7$ in vertex form and state its axis of symmetry. 21. Write $-2x^2+8x-5$ in vertex form and state its maximum value. 22. Expand $(x+4)^2-11$ into standard form and solve the corresponding equation. 23. Write $2(x-3)(x+5)$ in standard and vertex forms. Verify that the midpoint of its roots is the vertex's $x$-coordinate. 24. A quadratic has roots $-2$ and $4$ and leading coefficient $1$. Write it in factored, standard, and vertex forms. ## Discriminant and reasoning 25. Classify the roots of $2x^2+3x+7=0$ without solving. 26. Find all real values of $k$ for which $x^2+kx+9=0$ has exactly one real solution. 27. Find all real values of $m$ for which $x^2-6x+m=0$ has two distinct real solutions. 28. Prove from the quadratic formula that the average of two real roots is $-b/(2a)$. 29. For $y=2(x-1)^2-8$, find the roots, convert to standard form, and show that its discriminant predicts the same intercept count. 30. Construct a quadratic equation with roots $2$ and $5$ and leading coefficient $3$. Verify the sum and product relations in @eq:vieta. 31. Diagnose the error: from $(x-4)^2=11$, a solution states $x=4+\sqrt{11}$. 32. Diagnose the error: $x^2+6x=16$ is rewritten as $(x+3)^2=16$. 33. Solve $x^2-6x+5=0$ by factoring and by completing the square. Explain what the second form reveals that the first does not. 34. Solve $x^2+4x-1=0$ by completing the square and by the quadratic formula. Show that the exact answers agree. 35. Solve $x^2-2px+p^2-9=0$ for $x$ in terms of the real parameter $p$. 36. A monic quadratic has roots $r$ and $s$. Prove that its equation is $x^2-(r+s)x+rs=0$. ## Models and synthesis 37. A rectangle has area $54$ square units and its length is $3$ units greater than its width. Find its dimensions and explain why one algebraic root is rejected. 38. The height of a ball is $h(t)=-16t^2+48t+5$ feet. Find its maximum height and the time at which it occurs. Find when it reaches the ground, giving an exact value and a decimal to the nearest hundredth. 39. A simplified stopping-distance model is $d=0.05v^2+0.4v$, where $d$ is feet and $v$ is miles per hour. Find the positive speed that gives a stopping distance of $60$ feet, to the nearest tenth. 40. A rectangular garden beside a wall uses $80$ feet of fencing on the other three sides. Find all dimensions that produce an area of $600$ square feet. 41. A quadratic has roots $3\pm\sqrt5$ and $y$-intercept $-8$. Determine its equation in standard form. 42. For a real quadratic with $a>0$ and two distinct real roots $r_1