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Quadratic Equations






Second-Degree Equations and the Threshold of Multiplicity

When the unknown is squared, equations acquire a new dimension of complexity. A quadratic equation can produce two solutions, one solution, or no real solutions at all — and a single quantity built from its coefficients determines which case applies before any solving begins. Three distinct methods handle the solving itself, each suited to different situations but all ultimately equivalent. Quadratic equations sit at the boundary between elementary algebra and the richer structure of polynomial equations, and the tools developed here — the discriminant, factoring, completing the square, and the quadratic formula — recur throughout mathematics.



Definition and Standard Form

A quadratic equation in one variable is any equation that can be written as

ax2+bx+c=0ax^2 + bx + c = 0


where aa, bb, and cc are real constants and a0a \neq 0. The requirement a0a \neq 0 is what makes the equation genuinely quadratic — if aa were zero, the x2x^2 term would vanish and the equation would reduce to a linear equation.

The word "quadratic" derives from the Latin word for square, reflecting the presence of x2x^2 as the highest-degree term. The coefficient aa is the leading coefficient, bb is the linear coefficient, and cc is the constant term. Every quadratic equation can be brought into standard form by rearranging: moving all terms to one side and ordering by descending powers of xx.

The equation x2=5x6x^2 = 5x - 6, for instance, is not in standard form but is still quadratic. Subtracting 5x5x and adding 66 to both sides yields x25x+6=0x^2 - 5x + 6 = 0, with a=1a = 1, b=5b = -5, and c=6c = 6. Recognition is the first step — any equation reducible to the form above, regardless of how it is initially presented, is quadratic and amenable to the methods on this page.

Solving by Factoring

When the quadratic expression ax2+bx+cax^2 + bx + c can be decomposed into a product of two first-degree factors, the equation reduces to a pair of linear equations through the zero-product property: if a product of two factors equals zero, at least one of the factors must be zero.

Consider x25x+6=0x^2 - 5x + 6 = 0. The left side factors as (x2)(x3)=0(x - 2)(x - 3) = 0. Setting each factor equal to zero gives x2=0x - 2 = 0 or x3=0x - 3 = 0, so x=2x = 2 or x=3x = 3. Both values satisfy the original equation.

When the leading coefficient is not 11, factoring requires more care. The equation 6x2+x2=06x^2 + x - 2 = 0 factors as (2x1)(3x+2)=0(2x - 1)(3x + 2) = 0, giving x=12x = \frac{1}{2} or x=23x = -\frac{2}{3}. Finding these factors involves searching for two binomials whose product reconstructs the original trinomial — a process that amounts to finding two numbers whose product is acac and whose sum is bb.

Factoring is efficient when it works, but it has a fundamental limitation: not every quadratic with rational coefficients factors into binomials with rational entries. The equation x22=0x^2 - 2 = 0 has solutions x=±2x = \pm\sqrt{2}, which are irrational, so no factorization over the rationals exists. A method that handles every quadratic without exception requires a different approach.

Solving by Completing the Square

Completing the square transforms any quadratic equation into a form where the solution can be extracted by taking a square root. The method works unconditionally — it does not depend on whether the quadratic factors neatly.

Starting from ax2+bx+c=0ax^2 + bx + c = 0, divide every term by aa to obtain x2+bax+ca=0x^2 + \frac{b}{a}x + \frac{c}{a} = 0. Move the constant to the right side:

x2+bax=cax^2 + \frac{b}{a}x = -\frac{c}{a}


The left side is almost a perfect square trinomial. Adding (b2a)2\left(\frac{b}{2a}\right)^2 to both sides completes it:

x2+bax+b24a2=b24a2cax^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} - \frac{c}{a}


The left side factors as a perfect square:

(x+b2a)2=b24ac4a2\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}


Taking the square root of both sides and isolating xx gives the two solutions. The expression under the square root, b24acb^2 - 4ac, determines whether those solutions are real or complex.

As a concrete example, consider x2+6x+2=0x^2 + 6x + 2 = 0. Move the constant: x2+6x=2x^2 + 6x = -2. Half of 66 is 33, and 32=93^2 = 9. Add 99 to both sides: x2+6x+9=7x^2 + 6x + 9 = 7. Factor: (x+3)2=7(x + 3)^2 = 7. Take roots: x+3=±7x + 3 = \pm\sqrt{7}. The solutions are x=3+7x = -3 + \sqrt{7} and x=37x = -3 - \sqrt{7}.

The Quadratic Formula

Applying the method of completing the square to the general equation ax2+bx+c=0ax^2 + bx + c = 0 — with letters in place of specific numbers — produces a universal formula:

x=b±b24ac2ax = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}


This is the quadratic formula. It accepts any real coefficients aa, bb, cc (with a0a \neq 0) and returns the solutions directly. The ±\pm symbol indicates that two values are generated: one using the positive square root and one using the negative.

For the equation 2x23x5=02x^2 - 3x - 5 = 0, the coefficients are a=2a = 2, b=3b = -3, c=5c = -5. Substituting:

x=(3)±(3)24(2)(5)2(2)=3±9+404=3±494=3±74x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(2)(-5)}}{2(2)} = \frac{3 \pm \sqrt{9 + 40}}{4} = \frac{3 \pm \sqrt{49}}{4} = \frac{3 \pm 7}{4}


The two solutions are x=104=52x = \frac{10}{4} = \frac{5}{2} and x=44=1x = \frac{-4}{4} = -1.

The formula is not a separate technique from completing the square — it is the result of completing the square once, in general form, so that the work never needs repeating. Every specific application of completing the square arrives at the same place the formula reaches in a single substitution.

The Discriminant

The expression Δ=b24ac\Delta = b^2 - 4ac appearing under the square root in the quadratic formula is called the discriminant. Its value determines the nature and number of solutions without requiring the full computation.

When Δ>0\Delta > 0, the square root is a positive real number, and the formula produces two distinct real solutions. The equation x25x+3=0x^2 - 5x + 3 = 0 has Δ=2512=13>0\Delta = 25 - 12 = 13 > 0, so two different real roots exist.

When Δ=0\Delta = 0, the square root vanishes, and the formula collapses to a single value x=b2ax = -\frac{b}{2a}. The quadratic has a repeated root — a double root — and the polynomial factors as a(xr)2a(x - r)^2 where r=b2ar = -\frac{b}{2a}. The equation x26x+9=0x^2 - 6x + 9 = 0 has Δ=3636=0\Delta = 36 - 36 = 0 and the single solution x=3x = 3.

When Δ<0\Delta < 0, no real number has a negative square, so the square root does not produce a real value. Over the real numbers, the equation has no solutions. Over the complex numbers, the formula yields two conjugate roots x=b±iΔ2ax = \frac{-b \pm i\sqrt{|\Delta|}}{2a}, where ii is the imaginary unit. The equation x2+x+1=0x^2 + x + 1 = 0 has Δ=14=3<0\Delta = 1 - 4 = -3 < 0 and the complex roots x=1±i32x = \frac{-1 \pm i\sqrt{3}}{2}.

The discriminant is a diagnostic tool: it classifies the equation's solution structure from the coefficients alone.

Vieta's Formulas

The French mathematician François Viète established a direct relationship between the roots of a quadratic equation and its coefficients. For the equation ax2+bx+c=0ax^2 + bx + c = 0 with roots x1x_1 and x2x_2:

x1+x2=bax_1 + x_2 = -\frac{b}{a}


x1x2=cax_1 \cdot x_2 = \frac{c}{a}


These identities follow from factoring the quadratic as a(xx1)(xx2)=0a(x - x_1)(x - x_2) = 0 and expanding: a[x2(x1+x2)x+x1x2]=0a[x^2 - (x_1 + x_2)x + x_1 x_2] = 0. Matching coefficients with ax2+bx+cax^2 + bx + c produces both formulas.

Vieta's formulas serve two purposes. First, they verify solutions: after solving x27x+10=0x^2 - 7x + 10 = 0 and obtaining x1=2x_1 = 2, x2=5x_2 = 5, check that 2+5=7=712 + 5 = 7 = -\frac{-7}{1} and 25=10=1012 \cdot 5 = 10 = \frac{10}{1}. Both match, confirming the answers.

Second, they construct equations from known roots. If a quadratic equation must have roots x1=3x_1 = -3 and x2=4x_2 = 4, then x1+x2=1x_1 + x_2 = 1 and x1x2=12x_1 \cdot x_2 = -12, so the equation is x2x12=0x^2 - x - 12 = 0 (taking a=1a = 1). The roots determine the equation uniquely up to a constant multiple.

Reducible Equations

Certain equations of degree higher than two can be converted into quadratic equations through substitution. The key is recognizing that the equation's structure mirrors the quadratic pattern au2+bu+c=0au^2 + bu + c = 0, with some expression playing the role of uu.

The most common case is the bi-quadratic equation ax4+bx2+c=0ax^4 + bx^2 + c = 0. Setting u=x2u = x^2 transforms it into au2+bu+c=0au^2 + bu + c = 0, a standard quadratic in uu. Solving for uu yields up to two values, and each positive value of uu produces two values of xx via x=±ux = \pm\sqrt{u}. A negative value of uu contributes no real solutions for xx.

For example, x45x2+4=0x^4 - 5x^2 + 4 = 0 becomes u25u+4=0u^2 - 5u + 4 = 0 with u=x2u = x^2. Factoring gives (u1)(u4)=0(u - 1)(u - 4) = 0, so u=1u = 1 or u=4u = 4. Back-substituting: x2=1x^2 = 1 gives x=±1x = \pm 1, and x2=4x^2 = 4 gives x=±2x = \pm 2. The original quartic has four real solutions.

The substitution pattern extends beyond powers of xx. Any equation of the form a[f(x)]2+b[f(x)]+c=0a[f(x)]^2 + b[f(x)] + c = 0 is quadratic in f(x)f(x). Setting u=f(x)u = f(x) reduces it, and back-substitution after solving requires solving f(x)=u1f(x) = u_1 and f(x)=u2f(x) = u_2 separately. Extraneous solutions can arise during back-substitution, so every candidate must be checked in the original equation.

Connection to Graphing

The solutions of ax2+bx+c=0ax^2 + bx + c = 0 correspond to the points where the parabola y=ax2+bx+cy = ax^2 + bx + c meets the xx-axis. At these points the output value is zero, which is precisely what the equation demands. The discriminant controls the geometry of this intersection.

When Δ>0\Delta > 0, the parabola crosses the xx-axis at two distinct points, and the equation has two real solutions. The horizontal distance between these crossing points is Δa\frac{\sqrt{\Delta}}{|a|}, so a larger discriminant means the roots are farther apart.

When Δ=0\Delta = 0, the parabola is tangent to the xx-axis — it touches at exactly one point and turns back. The vertex of the parabola sits on the axis, and the double root x=b2ax = -\frac{b}{2a} is the xx-coordinate of that vertex.

When Δ<0\Delta < 0, the parabola lies entirely above the xx-axis (if a>0a > 0) or entirely below it (if a<0a < 0). It never reaches the axis, so no real intersection exists and the equation has no real solutions. The vertex is the closest the parabola comes to the axis, and the gap between them reflects how negative the discriminant is.

The sign of aa determines the parabola's orientation — opening upward when a>0a > 0 and downward when a<0a < 0 — but it does not affect the number of solutions. That role belongs exclusively to the discriminant.