Complex Numbers and Quadratic Equations · TOPIC LESSON
Polynomial Roots
Roots, multiplicities and signs, with a full two-variable quadratic-form extension
JEE Main · JEE Advanced100 min study guide13 indexed sections
BY THE END, YOU CAN
Use factor and remainder theorems, multiplicity and coefficient-root relations.
Transform root expressions without unnecessarily solving every root.
Analyze real quadratic roots and signs with discriminant, vertex and domain conditions.
Classify genuine homogeneous quadratic forms in two variables and connect their zeros to real line factors.
BEFORE YOU START
Polynomial algebra, quadratic equations and complex conjugates.
Intervals, inequalities and elementary matrix notation.
Original educational lesson. Work through the examples before revealing the practice answers. Check the current official syllabus separately; historical question frequency does not establish eligibility or predict future questions.
CONNECT THE CONCEPT TO THE EXAM
Selected exam questions & worked solutions
These selected examples are separate from the original practice below. Each identifies its exam, year and wording. They do not represent every paper or every selected year.
No source-verified exam questions have been published for this lesson yet. The worked examples and practice below are original teaching material, not past-exam questions.
01 / UNDERSTAND & APPLY
1. A root is an input where a polynomial vanishes
A polynomial P(x)=aₙxⁿ+…+a₁x+a₀ has degree n when aₙ≠0. A root or zero is a number r with P(r)=0. The zero polynomial is exceptional: every input is a root and it has no ordinary finite degree. When a parameter can make the leading coefficient vanish, separate that case before using degree-dependent conclusions.
Polynomial division by x−r gives P(x)=(x−r)Q(x)+P(r), because the constant remainder is found by substituting x=r. Thus r is a root exactly when x−r is a factor. More generally, the remainder after division by x−a is P(a), even when a is not a root.
A nonzero polynomial of degree n has at most n distinct roots. Over the complex numbers it has exactly n roots counted with multiplicity, by the fundamental theorem of algebra. These are different counts: (x−2)³ has one distinct root but three roots when multiplicity is counted.
If a polynomial has real coefficients, nonreal roots occur in conjugate pairs. Conjugating P(z)=0 gives P(conjugate z)=0 because the coefficients are unchanged. Rational or integer coefficient assumptions are stronger and support additional root tests; do not use them for arbitrary real-coefficient polynomials.
ORIGINAL WORKED EXAMPLE
Factor, count multiplicity and read the sign change
Factor P(x)=x³−3x²+4 and classify its real roots.
P(2)=8−12+4=0, so x−2 is a factor. Division gives x²−x−2=(x−2)(x+1).
Thus P(x)=(x−2)²(x+1). The distinct real roots are 2 and −1, with multiplicities 2 and 1.
Near x=2, the squared factor is nonnegative and x+1 is positive, so the polynomial has the same positive sign on both sides of 2.
Near x=−1, the squared factor is positive while x+1 changes sign, so the graph crosses the axis there. Multiplicity affects local sign, not merely the root list.
Answer P(x)=(x−2)²(x+1); roots 2 (double) and −1 (simple).
02 / UNDERSTAND & APPLY
2. Candidate tests and derivative tests have different jobs
For an integer-coefficient polynomial with nonzero constant term, a rational root p/q in lowest terms must have p dividing the constant coefficient and q dividing the leading coefficient. This gives candidates, not a guarantee that any candidate is a root. Substitute or divide to verify each one.
A root r has multiplicity m when P(x)=(x−r)ᵐQ(x) with Q(r)≠0. For real coefficients, Q keeps a fixed nonzero sign sufficiently near r. An odd m makes P change sign across r; an even m does not. An even-multiplicity root can therefore be missed by a sign-change-only search.
A repeated root is a common root of P and P′. More precisely, multiplicity m means P(r)=P′(r)=…=P^(m−1)(r)=0 and P^(m)(r)≠0 for a nonzero polynomial. Derivatives provide an algebraic repetition test, but P′(r)=0 alone does not mean P(r)=0.
When factoring a parameterized polynomial, preserve every factor and any parameter value used in division. A factor that is assumed nonzero can remove a special case. The correct strategy is to split the exceptional case, then solve the generic one with justified divisions.
03 / UNDERSTAND & APPLY
3. Coefficients encode symmetric information about roots
For ax²+bx+c with a≠0 and roots α,β counted with multiplicity, α+β=−b/a and αβ=c/a. These relations follow by expanding a(x−α)(x−β) and matching coefficients. They remain valid for complex roots, although order comparisons require real roots.
For a monic cubic x³−s₁x²+s₂x−s₃, the three roots satisfy α+β+γ=s₁, αβ+βγ+γα=s₂ and αβγ=s₃. The alternating signs come from choosing negative root terms during expansion, not from an arbitrary convention.
Symmetric expressions can often be found without individual roots. For two roots, α²+β²=(α+β)²−2αβ and α³+β³=(α+β)³−3αβ(α+β). Reciprocal sums require nonzero roots: 1/α+1/β=(α+β)/(αβ). Check the product before dividing.
Power sums also satisfy recurrences from the root equation. If each root obeys r²=Sr−P, then pₖ=αᵏ+βᵏ satisfies pₖ=Spₖ₋₁−Ppₖ₋₂ for k≥2, starting from p₀=2 and p₁=S. This computes high symmetric powers with little algebra.
ORIGINAL WORKED EXAMPLE
Find transformed root data without solving the quadratic
Let α,β be roots of x²−5x+3=0. Find α²+β² and a monic polynomial with roots 1/α,1/β.
Vieta gives α+β=5 and αβ=3. The nonzero product ensures both reciprocal roots exist.
The square sum is 5²−2·3=19.
The reciprocal-root sum is 5/3 and product is 1/3. Their monic polynomial is t²−(5/3)t+1/3.
Multiplying by 3 gives an equivalent equation 3t²−5t+1=0, but the requested monic polynomial retains leading coefficient 1.
4. Translate the root transformation into a substitution
If a new root is t=α+k, replace the old variable x by t−k in P(x). If t=cα with c≠0, replace x by t/c and clear nonzero constant denominators. This produces a polynomial with the transformed roots, retaining their multiplicities.
For reciprocal roots t=1/α, use tⁿP(1/t), provided P has no zero root. If the constant coefficient is zero, reciprocal transformation is undefined for that root and the degree can drop; remove or treat the zero-root factor explicitly.
For a nonlinear transformation such as t=α², different original roots can map to the same new value. Constructing the sum and product may be simpler than eliminating variables. State whether repeated transformed values are counted with multiplicity.
A common root of two polynomials can be found by eliminating their highest-degree terms or using the Euclidean algorithm. A polynomial combination vanishing at a common root supplies a necessary condition; substitute the candidate into both originals to confirm it.
ORIGINAL WORKED EXAMPLE
A shift and a repeated transformed value
For roots α,β of x²−2x−3=0, form equations whose roots are α+2,β+2 and α²,β².
For the shift t=x+2, substitute x=t−2: (t−2)²−2(t−2)−3=t²−6t+5.
Thus the shifted roots satisfy t²−6t+5=0. The original roots 3 and −1 become 5 and 1, a direct check.
For squares, α+β=2 and αβ=−3. The new sum is 2²−2(−3)=10 and the new product is (−3)²=9.
Therefore the squaring transformation gives t²−10t+9=0, whose roots are 9 and 1. In other problems opposite original roots would yield the same squared value twice.
5. The discriminant counts real roots; the leading coefficient sets the sign
For q(x)=ax²+bx+c with a≠0, complete the square: q(x)=a(x+b/(2a))²−Δ/(4a), where Δ=b²−4ac. The vertex lies at x=−b/(2a), and its value is −Δ/(4a). The discriminant is positive for two distinct real roots, zero for a repeated real root and negative for no real roots.
If a>0 and Δ<0, q is positive for every real x; if a<0 and Δ<0, it is negative for every real x. At Δ=0 it has the sign of a away from its repeated root but equals zero there. Strict positivity therefore requires Δ<0, whereas nonnegativity with a>0 allows Δ=0.
When Δ>0, order the roots r₁<r₂. The polynomial has the sign of a outside [r₁,r₂] and the opposite sign inside. Include or exclude the roots according to whether the requested inequality permits equality. A rational inequality also excludes denominator roots regardless of the inequality symbol.
For two real roots to be positive, their sum must be positive and product positive; the reality condition Δ≥0 must also be imposed. Sum and product alone can describe a pair of nonreal conjugates, so they do not replace the discriminant.
ORIGINAL WORKED EXAMPLE
A strict global sign condition excludes the boundary
Find all real m such that x²−2mx+m+2>0 for every real x.
The leading coefficient is 1>0, so strict positivity for every x is equivalent to a negative discriminant.
The discriminant is 4m²−4(m+2)=4(m−2)(m+1). Thus require (m−2)(m+1)<0.
The parameter lies between the roots: −1<m<2.
At m=−1 or m=2 the quadratic is a perfect square with a zero, so it is nonnegative but not strictly positive. Those boundaries must be excluded.
Answer m∈(−1,2).
06 / UNDERSTAND & APPLY
6. A root-location claim needs more than a discriminant
For a continuous polynomial, opposite endpoint signs guarantee at least one root between the endpoints. They do not give uniqueness by themselves, and they do not detect an even-multiplicity root where the sign does not change. Combine signs with degree, monotonicity or the vertex when counting roots.
For a quadratic opening upward, exactly one simple root on each side of a point t is equivalent to q(t)<0. For both distinct roots to lie in an interval, conditions must control reality, vertex position and endpoint signs. Applying only endpoint positivity can incorrectly place both roots outside the interval.
A useful method is to translate x=t+u when comparing roots with t. The new quadratic has roots α−t and β−t. Requiring both original roots above t becomes a positivity condition on the transformed roots, which can be checked using their sum, product and reality.
When the leading coefficient depends on a parameter, isolate values where it vanishes before applying quadratic criteria. The remaining equation may be linear, identically zero or inconsistent. A discriminant formula derived under a≠0 cannot classify those cases by itself.
ORIGINAL WORKED EXAMPLE
Locate both roots beyond a threshold
For x²−6x+k=0, find k such that both real roots are strictly greater than 1.
Set x=u+1. The transformed equation is u²−4u+(k−5)=0, whose roots must both be positive.
The transformed root sum is 4>0. A positive product requires k−5>0, so k>5.
For 5<k≤9, both real transformed roots are positive. At k=9 they coincide at u=2, still strictly positive. At k=5 one root is u=0 and fails the strict threshold.
Answer 5<k≤9.
07 / UNDERSTAND & APPLY
7. A genuine quadratic form has two variables and degree-two homogeneity
A real binary quadratic form is Q(x,y)=ax²+2hxy+by². Every term has total degree two, so Q(tx,ty)=t²Q(x,y) for every real t. It has no linear or constant terms. It is different from a one-variable quadratic equation, though one-variable root methods help analyze its directions.
The symmetric matrix M=[[a,h],[h,b]] represents the form as [x y]M[x y]ᵀ. The off-diagonal entry is h, half the coefficient of xy. Confusing h with 2h changes the determinant and every resulting sign test.
Positive definite means Q(x,y)>0 for every nonzero vector (x,y); both coordinates need not be nonzero. Negative definite similarly means strict negativity for every nonzero vector. Positive semidefinite means Q≥0 everywhere, and negative semidefinite means Q≤0 everywhere. Indefinite means the form takes both positive and negative values.
The zero vector always gives Q(0,0)=0, so strict definiteness deliberately excludes it. A semidefinite form may also vanish along a whole nonzero direction. The zero form is both positive and negative semidefinite but is neither strictly definite nor indefinite.
08 / UNDERSTAND & APPLY
8. Completing squares proves the classification
When a≠0, Q=a(x+(h/a)y)²+((ab−h²)/a)y². Let D=ab−h². If a>0 and D>0, both square coefficients are positive and they cannot both vanish unless x=y=0, so Q is positive definite. Conversely, evaluating at (1,0) and (−h/a,1) proves that these conditions are necessary.
Negative definiteness requires a<0 and D>0 by the same decomposition. If D<0, the two coefficients have opposite signs when a≠0, so suitable directions give opposite signs. If a=0 and D<0, then h≠0 and the cross term similarly produces both signs. Thus D<0 characterizes an indefinite binary form.
Positive semidefiniteness requires a≥0, b≥0 and D≥0. These conditions also handle a=0: D=−h²≥0 forces h=0, leaving by²≥0. Negative semidefiniteness requires a≤0,b≤0,D≥0. Merely replacing strict inequalities in the positive-definite leading-minor test can miss the necessary condition on b when a=0.
When D=0 and the form is nonzero, it is a signed square of a linear expression and is semidefinite. Its zero set contains a full line. This is the boundary between definite and indefinite behavior, not a case to discard as an algebraic nuisance.
ORIGINAL WORKED EXAMPLE
Classify a parameterized two-variable form including boundaries
Classify Qₖ(x,y)=x²+2kxy+4y² for every real k.
Here a=1,h=k,b=4, so D=4−k². Completing the square gives Qₖ=(x+ky)²+(4−k²)y².
For |k|<2, both coefficients are positive, making the form positive definite.
For k=2 or k=−2, Qₖ=(x±2y)². It is positive semidefinite but not definite because nonzero points on x=∓2y give zero.
For |k|>2, Qₖ(1,0)=1>0 while Qₖ(−k,1)=4−k²<0. These explicit vectors prove indefiniteness.
Answer Positive definite for |k|<2; positive semidefinite but not definite for |k|=2; indefinite for |k|>2.
09 / UNDERSTAND & APPLY
9. The zero form equation describes directions through the origin
The equation Q(x,y)=0 is homogeneous, so any nonzero solution generates a whole line through the origin by scaling. For y≠0, set t=x/y and divide by y² to obtain at²+2ht+b=0. Its discriminant is 4(h²−ab)=−4D.
If D<0, the binary form factors into two distinct real linear factors and its zero set is two distinct lines. If D=0 and the form is nonzero, the two factors coincide up to scaling and the zero set is one repeated line. If D>0, the form is definite and has no nonzero real zero, so its real zero set is only the origin.
Dividing by y² can miss the direction y=0, which must be checked separately. If a=0, Q=y(2hx+by), so y=0 is immediately one factor. This direct factorization is safer than pretending the directional equation remains quadratic when its leading coefficient vanishes.
The sign regions of an indefinite form can be read from its two line factors, just as a one-variable factored polynomial has sign intervals. Here the lines divide the plane into sectors. The product of the signed linear factors determines which sectors have positive or negative Q.
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Homogeneous forms scale quadratically: q(tx,ty)=t²q(x,y).
ORIGINAL WORKED EXAMPLE
Factor a homogeneous quadratic and test its signs
Analyze Q(x,y)=2x²+xy−3y².
The form factors as (2x+3y)(x−y). Its zero set is the union of x=y and 2x+3y=0.
In the standard notation 2h=1, so h=1/2. Thus D=2(−3)−(1/2)²=−25/4<0.
The negative determinant predicts indefiniteness and two distinct real line factors, agreeing with the factorization.
Explicitly Q(1,0)=2>0 and Q(0,1)=−3<0. These vectors also verify the sign classification without relying solely on a determinant rule.
Answer Q=(2x+3y)(x−y); two real zero lines; the form is indefinite.
10 / UNDERSTAND & APPLY
10. A form can measure directional size
A positive-definite form behaves like a squared length after an invertible linear change of coordinates, but it need not equal x²+y² in the original axes. Its positive level sets are ellipses centered at the origin. A negative-definite form has analogous negative levels; an indefinite form has hyperbola-type nonzero levels and two zero lines.
For a real symmetric matrix, an orthogonal change of coordinates diagonalizes the form with real eigenvalues. Positivity of both eigenvalues means positive definiteness, opposite signs mean indefiniteness, and a zero eigenvalue supplies a direction with zero value. In two variables the product of the eigenvalues is D and their sum is a+b.
One can often obtain sharp bounds without computing eigenvectors formally. Rewrite the form as a combination of (x+y)² and (x−y)², then impose x²+y²=1. The equality directions reveal where a minimum or maximum occurs and provide a direct check on the algebra.
This section does not identify every quadratic polynomial in two variables as a homogeneous form. Linear terms translate the center and constants change level sets; degeneracy of a full conic needs additional information. The form describes the degree-two part, which is only one part of that broader classification.
ORIGINAL WORKED EXAMPLE
Find exact extrema of a form on the unit circle
Find the minimum and maximum of Q=3x²+2xy+3y² subject to x²+y²=1.
Using the constraint, Q=3+2xy. Since (x−y)²≥0 and (x+y)²≥0, one has −1≤2xy≤1.
Therefore 2≤Q≤4.
The minimum 2 occurs when x=−y and x²+y²=1, giving (x,y)=±(1/√2,−1/√2).
The maximum 4 occurs when x=y, giving ±(1/√2,1/√2). Equivalently Q=2(x²+y²)+(x+y)², and its matrix eigenvalues are 2 and 4.
Answer Minimum 2 along x=−y; maximum 4 along x=y, with the stated unit vectors.
11 / UNDERSTAND & APPLY
11. Original practice: roots and coefficients
Check leading coefficients, multiplicities and nonzero-root requirements before applying formulas.
ORIGINAL PRACTICE · 1
Find the remainder when x³+2x−5 is divided by x−2.
Reveal answer & reasoning
7.
The remainder theorem gives P(2)=8+4−5=7.
ORIGINAL PRACTICE · 2
For roots α,β of 2x²−7x+3=0, find α+β and αβ.
Reveal answer & reasoning
7/2 and 3/2.
Compare coefficients with 2(x−α)(x−β), using the leading coefficient 2.
ORIGINAL PRACTICE · 3
For roots of x²−4x+1=0, find α³+β³.
Reveal answer & reasoning
52.
The sum is 4 and product 1, so the cube sum is 4³−3·1·4=64−12=52.
ORIGINAL PRACTICE · 4
How many distinct real roots does (x+1)⁴(x−3)² have, and where does its sign change?
Reveal answer & reasoning
Two distinct roots; no sign change at either.
The roots are −1 and 3, both of even multiplicity. Away from them the product of even powers is positive.
ORIGINAL PRACTICE · 5
Find m such that x²+mx+9 has a repeated real root.
Reveal answer & reasoning
m=±6.
The discriminant m²−36 must equal zero. Both values keep a genuine quadratic.
ORIGINAL PRACTICE · 6
Solve x²−5x+6≤0.
Reveal answer & reasoning
x∈[2,3].
Factor as (x−2)(x−3). The upward quadratic is nonpositive between its roots, including both endpoints.
ORIGINAL PRACTICE · 7
Can a real-coefficient cubic have exactly one nonreal root counted with multiplicity?
Reveal answer & reasoning
No.
Nonreal roots occur in conjugate pairs, so the number of nonreal roots counted with multiplicity is even.
ORIGINAL PRACTICE · 8
For kx²+x−1=0, what happens at k=0?
Reveal answer & reasoning
It becomes the linear equation x−1=0, with root 1.
The quadratic leading coefficient vanishes. Quadratic root-count criteria assuming k≠0 cannot be applied unchanged.
12 / UNDERSTAND & APPLY
12. Original practice: actual two-variable forms
For strict definiteness, test every nonzero vector, including points on the coordinate axes. When a determinant test gives a boundary case, exhibit its zero direction.
ORIGINAL PRACTICE · 1
Classify Q=x²+6xy+10y².
Reveal answer & reasoning
Positive definite.
Here h=3 and D=10−9=1>0 with a=1>0. Completing squares gives (x+3y)²+y², positive except at the origin.
ORIGINAL PRACTICE · 2
Classify Q=−2x²+4xy−3y².
Reveal answer & reasoning
Negative definite.
Here a=−2,h=2,b=−3 and D=6−4=2>0. Equivalently Q=−2(x−y)²−y².
ORIGINAL PRACTICE · 3
Classify Q=4x²−4xy+y² and find its zero set.
Reveal answer & reasoning
Positive semidefinite, not definite; zero line y=2x.
The form is (2x−y)². It is nonnegative but vanishes at every point on that nonzero line.
ORIGINAL PRACTICE · 4
Classify Q=2xy.
Reveal answer & reasoning
Indefinite.
It is positive at (1,1) and negative at (1,−1). Its matrix determinant is −1.
ORIGINAL PRACTICE · 5
Classify Q=−y².
Reveal answer & reasoning
Negative semidefinite, not negative definite.
It is never positive, but every vector (x,0), including nonzero ones, gives zero.
ORIGINAL PRACTICE · 6
For Q=x²+2hxy+9y², find h for positive definiteness.
Reveal answer & reasoning
−3<h<3.
The leading coefficient is positive and D=9−h² must be strictly positive. At the endpoints the form is only semidefinite.
ORIGINAL PRACTICE · 7
Factor x²−5xy+6y² and describe the zero directions.
Reveal answer & reasoning
(x−2y)(x−3y); lines x=2y and x=3y.
The directional quadratic t²−5t+6 has roots 2 and 3. The y=0 direction gives only the origin here, so no additional line is lost.
ORIGINAL PRACTICE · 8
Find the least value of 5x²−4xy+5y² when x²+y²=1.
Reveal answer & reasoning
3.
The expression is 5−4xy. Since xy≤1/2, its minimum is 3, attained at x=y=±1/√2.
13 / UNDERSTAND & APPLY
13. Keep the one-variable and two-variable questions distinct
For a polynomial, track degree, roots and multiplicity. For a quadratic equation, track discriminant and the signs of the root sum and product. For a homogeneous quadratic form, track every nonzero direction in the plane and distinguish strict from weak sign conditions.
The bridge between the subjects is useful but conditional. Dividing Q(x,y) by y² produces a directional polynomial only when y≠0. Its missing axis direction must be checked. A coefficient-root transformation likewise needs every denominator and every zero-root exception to be controlled.
The common mistakes are ignoring a vanishing leading coefficient, confusing distinct roots with multiplicities, assuming positive sum and product imply real roots, using h as the full xy coefficient, and replacing positive definiteness by a test that checks only x and y both nonzero.
Finish with a structural check: substitute candidate roots, expand factors, compare coefficient sums, or exhibit vectors where a form is positive, negative or zero. These direct checks expose errors that a memorized discriminant slogan can conceal.
KEEP THE BIG IDEAS
Quick revision
Factor and remainder theorems connect values with polynomial divisibility.
Vieta relations encode symmetric root information without solving each root.
Real-root and sign questions require degree and discriminant conditions.
A binary quadratic form is genuinely homogeneous in two variables.
Completing squares and D=ab−h² classify definiteness and real zero directions.
The teaching notes, worked explanations and original practice are written for this lesson. Separately labeled selected exam questions identify their exam, year, source and whether wording is paraphrased or an exact excerpt. Educational references do not certify current exam coverage.
Author-hosted educational reference inspected for definitions and two-variable definiteness conditions. The lesson independently derives the tests and supplies original factorization, parameter and unit-circle problems.