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. Several 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, the quadratic formula, and Vieta's formulas — recur throughout mathematics.
Linear equations are solved by undoing operations one at a time.
In 5x−6=4, add 6, then divide by 5: x=2. Each step peels one operation off x, and the answer is a single number.
x² = 5x − 6 cannot be peeled that way — and it has two solutions, not one.
The unknown appears twice, once squared and once not. Moving terms around gives x2−5x+6=0, but no single "undo" step frees x. Trying values shows why one answer could never be enough: the left side hits zero twice.
x
x² − 5x + 6
is it 0?
0
6
no
1
2
no
2
0
yes
3
0
yes
4
2
no
5
6
no
The definition
A quadratic equation is an equation in one unknown that can be rearranged so the highest power of the unknown is 2 — the form ax2+bx+c=0 with a=0.
A quadratic equation in one variable is any equation that can be written as
Definition and Standard Form
ax2+bx+c=0
where a, b, and c are real constants and a=0. The requirement a=0 is what makes the equation genuinely quadratic — if a were zero, the x2 term would vanish and the equation would reduce to a linear equation.
So x2=5x−6 is quadratic, with a=1, b=−5, c=6 — the next step up from linear, the way a square's area is one step up from a side's length.
It is an equation recognized by its form, however it is first written.
The word "quadratic" derives from the Latin word for square, reflecting the presence of x2 as the highest-degree term. The coefficient a is the leading coefficient, b is the linear coefficient, and c 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 x.
The equation x2=5x−6, for instance, is not in standard form but is still quadratic. Subtracting 5x and adding 6 to both sides yields x2−5x+6=0, with a=1, b=−5, and c=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.
x² − 2x − 3 = 5: a quadratic not yet in standard form
The left side is drawn as a parabola and the right side 5 as a level above the axis; the solutions are where the two meet. Moving the 5 across gives x² − 2x − 8 = 0, which lowers the parabola by 5 so the same solutions become its crossings of the axis, the standard-form picture. Shift the level yourself on the equation visual explorer.
Bringing an equation to standard form never changes its solutions, only how they are read.
Nothing is lost: drop the x² term and the linear equation is back.
With a=0, the form ax2+bx+c=0 becomes bx+c=0, solved by the old undoing steps. Every linear technique still applies inside the new methods — factoring and the quadratic formula both end in small linear equations.
What is lost is the one-step answer: a quadratic can have two solutions, one, or none among the real numbers.
Undoing operations no longer works on its own, so new methods are needed — factoring, completing the square and the quadratic formula — and the number of solutions has to be found, not assumed. The section on solving by factoring starts with the most direct one.
The rest of the page covers the notation, solving by factoring, by completing the square and by the quadratic formula, the discriminant, Vieta's formulas and reducible equations.
Quadratic Notation
Notation
Quadratic Notation
The contract behind a, b, c, the two letters that both mean discriminant, and what a lowered index does that a raised one does not.
Standard form — descending powers, everything on the left
A notational contract, not just a tidy habit: a, b, c have meaning only relative to this form. The formula and the discriminant read their inputs from it — feed them coefficients from an unrearranged equation and every sign is suspect.
CasesFrom x2=5x−6 the contract reads b=−5, not 5 — the sign travels with the term during the rearrangement in Definition and Standard Form above. The function f(x)=ax2+bx+c is a different object: a parabola; the =0 asks only where it crosses.
Also writtenMonic form x2+px+q=0 — after dividing by a; Vieta's formulas are cleanest there, as p and q pick up the sum and product directly.
Do not confuseA claim about x. Standard form asserts nothing by itself — it is a position for the pieces, the notational equivalent of lining up digits before long division.
Δ=b2−4ac
Delta — the discriminant
One number, two letter traditions: Δ in most international and university texts — this page's choice, in The Discriminant below — and capital D in Russian, German, and much school-level writing. Same definition, same verdicts.
CasesThe sign gives the verdict: Δ>0 two real roots, Δ=0 a double root, Δ<0 a conjugate pair. Beyond the sign, its size only sets distances on the graph, as Connection to Graphing below shows.
Also writtenD=b2−4ac — interchangeable; pick whichever your course uses and stay with it.
Do not confuseThe other deltas. Δx is an increment and Δ alone often means “change in” — the discriminant Δ is a single fixed number computed from coefficients, related to neither.
x1,x2
x one and x two — the two roots
Lowered indices enumerate: x1 and x2 are two different numbers sharing a letter. Position is everything — lowered counts, raised operates: x2 is the second root, x2 is a square.
CasesA double root is written x1=x2 — two labels, one value. Vieta's formulas in Vieta's Formulas below treat the pair symmetrically, which is why the labels never need to say which root is which.
Also writtenx′, x′′ in some older European texts — a prime-based enumeration that collides with derivative notation and has largely died out for exactly that reason.
Do not confuseSequence indexing. In sequences the subscript an runs over infinitely many positions; here it is a finite enumeration — two names, fixed in advance, nothing running.
Polynomials, Equations and Inequalities: Notation at a Glance
The marks from the notation cards across the polynomial, equation and inequality pages: how each is read and what it means. Every group links to the full card on its page.
Polynomials, Equations and Inequalities: Notation at a Glance·/algebra/polynomials/rootsLearn Math Class
Solving by Factoring
RULEZero-product property
When the quadratic expression ax2+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.
Example
Consider x2−5x+6=0. The left side factors as (x−2)(x−3)=0. Setting each factor equal to zero gives x−2=0 or x−3=0, so x=2 or x=3. Both values satisfy the original equation.
Pitfall: Factoring before the right side is zero
Setting each factor of x(x−1)=6 equal to 6 gives x=6 or x=7, but 6⋅5=30. Moving the 6 first gives x2−x−6=(x−3)(x+2)=0, so x=3 or x=−2.
When the leading coefficient is not 1, factoring requires more care. The equation 6x2+x−2=0 factors as (2x−1)(3x+2)=0, giving x=21 or x=−32. 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 ac and whose sum is b.
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 x2−2=0 has solutions x=±2, which are irrational, so no factorization over the rationals exists. A method that handles every quadratic without exception requires a different approach.
Fact: Three odd coefficients, no rational roots
If a, b and c are all odd, ax2+bx+c=0 has no rational root. An odd b2 leaves remainder 1 on division by 8 and 4ac leaves 4, so b2−4ac leaves 5, a remainder no perfect square has.
Whether a search for factors can succeed is decided before it starts. With integer coefficients, a quadratic factors over the rationals exactly when b2−4ac, the quantity under the square root in the formula derived below, is a perfect square. For x2−5x+6 it is 25−24=1, and the factors exist; for x2−2 it is 8, and no factors with rational entries exist.
Factoring turns one quadratic into two linear equations
The section's x² − 5x + 6 = (x − 2)(x − 3). The two factor lines cross the axis at x = 2 and x = 3, and the parabola, their product, is zero at exactly those inputs and nowhere else: a product vanishes only when one of its factors does.
The zero-product property is why the right side must be 0 before factoring.
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.
Taking the square root of both sides and isolating x gives the two solutions. The expression under the square root, b2−4ac, determines whether those solutions are real or complex.
Example
As a concrete example, consider x2+6x+2=0. Move the constant: x2+6x=−2. Half of 6 is 3, and 32=9. Add 9 to both sides: x2+6x+9=7. Factor: (x+3)2=7. Take roots: x+3=±7. The solutions are x=−3+7 and x=−3−7.
x²3x3x54x +3x +3(x + 3)² − 4 = 0
x² + 6x + 5 = 0 becomes (x + 3)² − 4 = 0
The x² block and the 6x strip, split into two 3x halves, form an incomplete square; adding the 3 × 3 corner completes it, and the constant 5 falls short of that 9 by 4, so the equation reads (x + 3)² = 4 and x + 3 = ±2. The method is this rearrangement of area. Step through the corner and the gap on the completing the square visualizer.
The quadratic formula in the next section is this construction carried out once with letters.
The Quadratic Formula
Applying the method of completing the square to the general equation ax2+bx+c=0 — with letters in place of specific numbers — produces a universal formula:
This is the quadratic formula. It accepts any real coefficients a, b, c (with a=0) and returns the solutions directly. The ± symbol indicates that two values are generated: one using the positive square root and one using the negative.
For the equation 2x2−3x−5=0, the coefficients are a=2, b=−3, c=−5. Substituting:
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 section's 2x² − 3x − 5 = 0. The term −b/2a = 3/4 is the parabola's axis of symmetry; √Δ/2a = √49/4 = 7/4 is how far each root sits from it. The ± sends one root left and one right: 3/4 − 7/4 = −1 and 3/4 + 7/4 = 5/2.
When the discriminant is 0 the half-width vanishes and the two roots merge on the axis.
Taking the formula apart: the ± sign, the discriminant under the root, and the 2a below.
completing the square once, in general form: put in a, b, c and read off both solutions
x=−b±√b2−4ac2a
1The solutions
2Centre line
3Two solutions
4Discriminant
5Divides everything above
1The solutionsx
What the formula returns: the roots of ax² + bx + c = 0, the points where the parabola meets the x-axis. Example: for 2x² − 3x − 5 = 0 the formula gives x = 5/2 and x = −1.
2Centre line−b
Divided by 2a, −b gives the vertex line x = −b/2a, halfway between the two solutions. When the discriminant is 0, the single solution sits exactly there. Example: b = −3, so −b = 3 and the centre is 3/4, halfway between 5/2 and −1.
3Two solutions±
One solution adds the square root, the other subtracts it, so the two solutions sit symmetrically around −b/2a. Example: (3 + 7)/4 = 5/2 and (3 − 7)/4 = −1.
4Discriminantb² − 4ac
The discriminant: positive gives two real solutions, zero gives one, negative gives none. Its square root sets the gap between them, √Δ / |a|. Example: (−3)² − 4(2)(−5) = 49, √49 = 7, gap 7/2.
5Divides everything above2a
The whole numerator goes over 2a, and the formula needs a ≠ 0. Example: 2a = 4, so all of 3 ± 7 is divided by 4. Dividing only the 7 gives 3 ± 7/4, which is wrong.
The part most often misread is the denominator: 2a divides −b as well as the square root. The discriminant alone tells how many real solutions to expect.
The Discriminant
The expression 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, the square root is a positive real number, and the formula produces two distinct real solutions. The equation x2−5x+3=0 has Δ=25−12=13>0, so two different real roots exist.
When Δ=0, the square root vanishes, and the formula collapses to a single value x=−2ab. The quadratic has a repeated root — a double root — and the polynomial factors as a(x−r)2 where r=−2ab. The equation x2−6x+9=0 has Δ=36−36=0 and the single solution x=3.
Pitfall: Losing the sign of a negative c
Computing −4ac for x2+2x−3=0 as −12 gives Δ=4−12=−8 and no real roots. With c=−3, −4ac=−4(1)(−3)=12, so Δ=16 and x=1 or x=−3.
When Δ<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=2a−b±i∣Δ∣, where i is the imaginary unit. The equation x2+x+1=0 has Δ=1−4=−3<0 and the complex roots x=2−1±i3.
The discriminant is a diagnostic tool: it classifies the equation's solution structure from the coefficients alone.
Discriminant Δ = b² − 4ac
# real roots
Type of roots
Example
Δ > 0
2
two distinct real roots
x² − 5x + 3 = 0 (Δ = 13)
Δ = 0
1
one repeated (double) real root
x² − 6x + 9 = 0 (Δ = 0)
Δ < 0
0
two complex conjugate roots
x² + x + 1 = 0 (Δ = −3)
x² + 4 = 0: the parabola never reaches the axis
The vertex sits at height 4 above the axis, so no real x makes the expression zero: the discriminant 0² − 4 · 1 · 4 = −16 is negative and there are no crossings to mark. Lower the parabola by 4 and the crossings merge into one at the vertex; lower it further and two appear. Watch the discriminant change sign on the equation visual explorer.
Positive, zero, negative: two crossings, one, none.
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=0 with roots x1 and x2:
These identities follow from factoring the quadratic as a(x−x1)(x−x2)=0 and expanding: a[x2−(x1+x2)x+x1x2]=0. Matching coefficients with ax2+bx+c produces both formulas.
Vieta's formulas serve three purposes. First, they verify solutions: after solving x2−7x+10=0 and obtaining x1=2, x2=5, check that 2+5=7=−1−7 and 2⋅5=10=110. Both match, confirming the answers.
Second, they construct equations from known roots.
Example
If a quadratic equation must have roots x1=−3 and x2=4, then x1+x2=1 and x1⋅x2=−12, so the equation is x2−x−12=0 (taking a=1).
COROLLARY
The roots determine the equation uniquely up to a constant multiple.
Fact: A number and its reciprocal share decimals
The golden ratio φ≈1.618 solves x2−x−1=0, whose roots multiply to −1 and add to 1. So the other root is −φ1, and φ−φ1=1 gives φ1≈0.618, with exactly the decimals of φ.
Third — and often overlooked — they can be used to solve quadratic equations directly when the roots are simple integers. Instead of factoring or applying the formula, the question becomes: which two numbers sum to −ab and multiply to ac? For x2−7x+10=0, this asks for two numbers summing to 7 and multiplying to 10 — namely 2 and 5. The roots are read off by inspection. This approach is fastest when applicable but limited: it works only when the roots are obvious enough to spot, which restricts it to quadratics with integer or simple rational solutions. For irrational or complex roots, the formula remains necessary. In practice, Vieta's inspection is essentially a special case of factoring — finding the factorization (x−x1)(x−x2) by going through the roots rather than the trinomial — but it is fast enough on simple cases to deserve recognition as a method in its own right.
Quantity
In terms of roots
In terms of coefficients
Use
Sum
x₁ + x₂
−b ⁄ a
solve by inspection, verify, construct from roots
Product
x₁ · x₂
c ⁄ a
solve by inspection, verify, construct from roots
The coefficients already hold the sum and product of the roots, so roots can be checked, built or guessed without solving.
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=0, with some expression playing the role of u.
The most common case is the bi-quadratic equation ax4+bx2+c=0. Setting u=x2 transforms it into au2+bu+c=0, a standard quadratic in u. Solving for u yields up to two values, and each positive value of u produces two values of x via x=±u. A negative value of u contributes no real solutions for x.
Example
For example, x4−5x2+4=0 becomes u2−5u+4=0 with u=x2. Factoring gives (u−1)(u−4)=0, so u=1 or u=4. Back-substituting: x2=1 gives x=±1, and x2=4 gives x=±2. The original quartic has four real solutions.
Pitfall: Stopping at the substitute variable
Giving u=4 and u=−1 as the solutions of x4−3x2−4=0 answers for u, not for x. Back-substituting, x2=4 gives x=±2 and x2=−1 gives no real x, so the real solutions are x=±2.
The substitution pattern extends beyond powers of x. Any equation of the form a[f(x)]2+b[f(x)]+c=0 is quadratic in f(x). Setting u=f(x) reduces it, and back-substitution after solving requires solving f(x)=u1 and f(x)=u2 separately. Extraneous solutions can arise during back-substitution, so every candidate must be checked in the original equation.
Original form
Substitute
Becomes
Recover x by
ax⁴ + bx² + c = 0 (bi-quadratic)
u = x²
au² + bu + c = 0
x = ±√u for each u ≥ 0
a[f(x)]² + b·f(x) + c = 0 (general)
u = f(x)
au² + bu + c = 0
solve f(x) = u₁ and f(x) = u₂; verify all candidates
An equation that is quadratic in some expression is solved for that expression first, then each value is turned back into x.
Connection to Graphing
RULE
The solutions of ax2+bx+c=0 correspond to the points where the parabola y=ax2+bx+c meets the x-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, the parabola crosses the x-axis at two distinct points, and the equation has two real solutions. The horizontal distance between these crossing points is ∣a∣Δ, so a larger discriminant means the roots are farther apart.
When Δ=0, the parabola is tangent to the x-axis — it touches at exactly one point and turns back. The vertex of the parabola sits on the axis, and the double root x=−2ab is the x-coordinate of that vertex.
When Δ<0, the parabola lies entirely above the x-axis (if a>0) or entirely below it (if a<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 is 4∣a∣∣Δ∣, so it grows with how negative the discriminant is.
The sign of a determines the parabola's orientation — opening upward when a>0 and downward when a<0 — but it does not affect the number of solutions. That role belongs exclusively to the discriminant.
Discriminant
x-axis intersection
Geometric behavior
Δ > 0
two distinct points
parabola crosses the axis; gap between roots is √Δ ⁄ |a|
Δ = 0
one point (tangent)
vertex sits on the axis at x = −b ⁄ 2a
Δ < 0
none
parabola entirely above (a > 0) or below (a < 0) the axis
x² − 4 = 0: the solutions are the x-intercepts
The two crossings of the parabola with the axis, at −2 and 2, are the two solutions of the equation: solving a quadratic and finding the x-intercepts of its graph are the same task. The vertex halfway between them sits on the axis of symmetry x = −b/2a. Drag the parabola and watch the intercepts move together on the equation visual explorer.
Every algebraic method on this page locates these two points without drawing the curve.
Summary of Solving Methods
Four methods cover every quadratic equation in practice. Two are universal — completing the square and the quadratic formula always work. Two depend on the equation's specific structure: factoring needs a clean rational decomposition, and Vieta's inspection needs roots simple enough to spot. The table below summarizes when to reach for each, whether it is universal, and what insight each one produces beyond the solution itself.
Method
When to use
Universal?
Notes
Factoring
when the quadratic factors over the rationals
✗
fastest when applicable; uses the zero-product property
Completing the square
always
✓
conceptual root; produces the quadratic formula in the general case
This summary collects the formulas of the page in one table, in the order the page builds them. Each row gives the name with a line on where the formula comes from and a link to its section, the formula itself, and what it gives.
It starts from the standard form and the square root property, then follows completing the square from the identity to the completed form of the general equation. The quadratic formula comes out of that last line, the discriminant is the part under its square root, and Vieta's formulas close the table with the sum and the product of the roots.
Read from top to bottom, the table is one derivation: completing the square on the standard form produces the quadratic formula. The discriminant then says how many real solutions to expect before computing them, and Vieta's formulas check the answer once it is found.
Common Pitfalls
Five mistakes account for most wrong answers to quadratic equations. Each one is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
The equation is already written as a product, but the product equals a number other than zero. The zero-product property is applied anyway, and each factor is set equal to that number.
Start
x(x−1)=6
Wrong turn
Error here
⟹x=6 or x−1=6
⟹x=6 or x=7
Right path
Move every term to one side first.
x2−x−6=0
⟹(x−3)(x+2)=0
⟹x=3 or x=−2
Why it happensThe left side already looks factored, so bringing the equation to standard form seems unnecessary. But the zero-product property works only because zero is special: a product is 0 only when one factor is 0, while a product of 6 can come from 2⋅3, 12⋅21 or infinitely many other pairs.
How to checkSubstitute each answer into the original equation. x=6 gives 6⋅5=30 and x=7 gives 7⋅6=42, neither equal to 6. The right answers give 3⋅2=6 and (−2)(−3)=6.
2Dropping the negative square root
After completing the square, the equation has the form (x+k)2=p with p>0, and only the positive square root of p is taken.
Start
(x−2)2=9
Wrong turn
Error here
⟹x−2=3
⟹x=5
Right path
Both 3 and −3 square to 9.
⟹x−2=±3
⟹x=5 or x=−1
Why it happensThe symbol 9 means only the positive root, 3, and taking the square root of both sides is read as applying that symbol. The equation asks for every number whose square is 9, and there are two.
How to checkWhen p>0, the equation (x+k)2=p always has two roots, placed symmetrically around x=−k. Here 5 and −1 lie 3 units on either side of 2, and (−1−2)2=9.
3Dividing only the root by 2a
The quadratic formula is copied with the fraction bar under the square root alone, so −b is left outside the division.
Start
For x2−6x+5=0, with a=1, b=−6, c=5 and Δ=36−20=16:
Wrong turn
Error here
x=6±216=6±2
⟹x=8 or x=4
Right path
The whole numerator −b±Δ is divided by 2a.
x=26±4
⟹x=5 or x=1
Why it happensThe formula is often written or remembered in one line, and the length of the fraction bar gets lost. On a calculator the same slip appears when the numerator is not put in brackets.
How to checkBy Vieta's formulas the roots must add up to −ab=6 and multiply to ac=5. The wrong pair gives 8+4=12 and 8⋅4=32; the right pair gives 5+1=6 and 5⋅1=5.
4Losing the sign of a negative c
The discriminant is computed with the minus sign in −4ac but without the sign of c itself, so a negative c is treated as positive.
Start
For x2+2x−3=0:
Wrong turn
Error here
Δ=22−4⋅1⋅3=4−12=−8<0
Which suggests the equation has no real solutions.
Right path
With c=−3,
Δ=22−4(1)(−3)=4+12=16>0
So there are two real roots, x=2−2±4, that is x=1 or x=−3.
Why it happensThe formula already contains a minus sign, and writing the coefficients without brackets hides the second one.
How to checkWhen a and c have opposite signs, −4ac is positive, so Δ=b2−4ac>0 and the equation always has two real roots. A negative discriminant in that case means a sign was lost. Factoring confirms the roots here: (x−1)(x+3)=x2+2x−3.
5Stopping at the substitute variable
In a reducible equation, the substitution u=x2 turns the problem into a quadratic in u, and the values of u are reported as the answer.
Start
x4−3x2−4=0
⟹u2−3u−4=0
⟹(u−4)(u+1)=0
Wrong turn
Error here
⟹u=4 or u=−1
Right path
Each value of u is turned back into an equation in x. From x2=4, x=2 or x=−2. From x2=−1 there is no real x. The real solutions are x=2 and x=−2.
Why it happensSolving the quadratic in u feels like the end of the problem, because it is the step where the familiar methods are used. The question, though, was asked about x.
How to checkSubstitute the answers into the original equation. x=2 gives 16−12−4=0. Read as values of x, the numbers 4 and −1 fail: x=4 gives 256−48−4=204 and x=−1 gives 1−3−4=−6.
Three quick checks between them catch all five slips. Substituting the answers into the original equation exposes the factoring done too early and the stop at the substitute variable. Vieta's formulas, with the roots adding up to −ab and multiplying to ac, expose the half-divided formula and the missing negative root. And when a and c have opposite signs the equation always has two real roots, so a negative discriminant there means a sign was lost.
Quadratic Equations FAQ
In x2=5x−6, is b equal to 5 or −5?
+
It is −5. The letters a, b and c mean nothing until the equation sits in standard form, so rearrange first: x2−5x+6=0. The sign travels with the term as it crosses, giving a=1, b=−5, c=6. Reading coefficients off an unrearranged equation makes every sign in the formula suspect.Read more →
Is the discriminant written Δ or D?
+
Both, and they mean the same thing. International and university texts favour Δ, while Russian, German and much school-level writing use capital D; the definition b2−4ac and the verdicts are identical. Pick whichever your course uses and stay consistent. Note that this Δ is unrelated to Δx, which means a change in x.Read more →
What is the difference between x2 and x2?
+
Position decides: a lowered index enumerates, a raised one operates. So x2 is the second root, a number in its own right, while x2 is x multiplied by itself. A double root is written x1=x2, meaning two labels pointing at one value, which is why the subscripts never have to say which root is which.Read more →
Can a quadratic with all odd coefficients have a rational root?
+
No. With integer coefficients, ax2+bx+c=0 has rational roots only when b2−4ac is a perfect square. If a, b and c are all odd, b2 leaves remainder 1 on division by 8 and 4ac leaves 4, so b2−4ac leaves 5, and no perfect square does. So 3x2+5x+1=0, for example, can only have irrational or complex roots.Read more →
Why must one side be zero before you factor?
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Because the zero-product property works only for zero: a product is 0 only when one of its factors is 0. A product of 6 can come from 2⋅3, 12⋅21 or infinitely many other pairs, so setting each factor of x(x−1)=6 equal to 6 gives x=6 or x=7, both wrong. Rearranged, x2−x−6=(x−3)(x+2)=0 gives x=3 or x=−2.Read more →
Does a quadratic always have real roots when a and c have opposite signs?
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Yes, always two distinct ones. Then ac<0, so −4ac is positive and Δ=b2−4ac is greater than b2, which is never negative. That holds whatever b is: x2+2x−3=0 has Δ=16 and the roots 1 and −3. A negative discriminant in that case means a sign was lost, usually the sign of c.Read more →
What does the size of the discriminant tell you?
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Distances on the graph; the sign alone gives the number of real roots. When Δ>0 the two roots are ∣a∣Δ apart: for 2x2−3x−5=0, Δ=49 and the roots −1 and 25 are 27 apart. When Δ<0 the vertex sits 4∣a∣∣Δ∣ from the axis: for x2+4=0, Δ=−16 and the vertex is 4 units above it.Read more →
