Common roots — common mistake

“Two intersecting roads that cross at exactly one junction satisfy both equations; the resultant condition guarantees a single common point.”

Two quadratics a x²+bx+c=0 and a' x²+b'x+c'=0 share a root iff (a c'−a' c)² = (a b'−a' b)(b c'−b' c). This follows from the resultant that eliminates x. The classic mistake is to equate coefficients term-by-term, which only works when the equations are identical, not when they merely share one root.

Memory trick: Common root ⇔ (ac'−a'c)² = (ab'−a'b)(bc'−b'c)

Sign of a quadratic

“A basketball hoop’s net is like the x-axis; a ball (quadratic) either stays above, dips below, or just grazes the rim at one point.”

For f(x)=ax²+bx+c, if D<0 the curve never meets the x-axis, so its sign is the sign of a everywhere. If D=0 it touches at the vertex, sign stays a on both sides. If D>0 the sign equals a outside the two distinct roots and opposite a between them. Trap: forgetting the D=0 case and assuming a sign change.

Memory trick: a·f(x)>0 → same sign as a; D=0 → touch, no flip

Sign of a quadratic — common mistake

“A seesaw tilts according to the heavier side; ignoring a is like forgetting the weight.”

The sign of a quadratic depends first on the leading coefficient a. If a>0 the parabola opens upward; if a<0 it opens downward. Ignoring a leads to wrong sign conclusions, especially when Δ>0.

Memory trick: Check a before roots.

Maximum or minimum

“A ball’s highest point occurs at time t = –b/2a, just like the vertex of a quadratic.”

The vertex of y = ax²+bx+c is at x = –b/2a; the corresponding y-value gives the maximum if a<0 or minimum if a>0. A common pitfall is stating only the x-coordinate without evaluating f(x).

Memory trick: Vertex x=–b/2a, then y=f(x).

Maximum or minimum — common mistake

“Guessing the roller-coaster’s peak without the design formula misses the vertex rule.”

Extrema of a quadratic are found analytically at x = –b/2a, not by trial. Forgetting to compute the y-value or using a guess can cost marks.

Memory trick: Never guess; use –b/2a.

Location of roots

“Imagine a road sign at x=k; the quadratic’s roots are cars; whether they pass before or after the sign depends on the sign of f(k) and the road’s slope a.”

Let f(x)=ax²+bx+c with D>0. If a·f(k)>0 then k lies outside the root interval; if a·f(k)<0 then k lies between the roots. Moreover, when k is to the right of the vertex (k>-b/2a) and a·f(k)>0, both roots are left of k; when k is left of the vertex, both roots are right of k. Trap: applying the rule when D≤0 or ignoring the vertex’s position.

Memory trick: Same sign with a → outside; opposite → between; compare k with vertex

Location of roots — common mistake

“A GPS showing distance (discriminant) tells you you’re reachable, but not whether the destination lies north or south of your current point.”

The discriminant D=b²-4ac only tells whether roots are real, equal or complex; it does not locate them relative to a number k. To decide the position, evaluate f(k) and the vertex x_v= -b/2a. Ignoring f(k) leads to the common error of assuming D>0 means roots lie on both sides of any chosen k.

Memory trick: D tells count, not location – add f(k) & vertex.

Polynomial remainder and factor theorem

“Testing a key (plugging a into f) instantly tells if the lock (x-a) fits without cutting the key.”

Remainder theorem: when f(x) is divided by (x-a) the remainder equals f(a). Hence f(a)=0 ⇔ (x-a) is a factor ⇔ a is a root. This holds for any degree polynomial. Trap: thinking the theorem applies only when the divisor is linear with leading coefficient 1; any (x-a) works.

Memory trick: Plug a → remainder.

Polynomial remainder and factor theorem — common mistake

“Using a hammer (long division) to open a bottle (check f(a)) is overkill; the key (substitution) does it instantly.”

Factor theorem gives the answer instantly: if f(a)=0 then (x-a) divides f(x). Performing long division wastes time and may introduce sign mistakes. The exam shortcut is to substitute first; only divide if f(a)≠0 and you need the quotient.

Memory trick: If f(a)=0, stop – factor found.

x = (-b +/- sqrt(b^2-4ac)) / 2a

“A GPS gives two routes (±) to the same spot; the quadratic formula lists both possible x-values.”

For ax²+bx+c=0 (a≠0) the roots are x = [-b ± √(b²-4ac)]/(2a). The formula is valid only after confirming D≥0 for real roots (or using complex √). A frequent slip is to forget the ± or to divide by a instead of 2a.

x = (-b +/- sqrt(b²-4ac)) / 2a

Memory trick: ±√D over 2a.

Sum of roots = -b/a, product = c/a

“Two siblings’ ages add to -b/a and multiply to c/a – that’s how Vieta links roots to coefficients.”

If α and β are the roots of ax²+bx+c=0, then α+β = –b/a and αβ = c/a, regardless of whether they are real or complex. These relations follow from expanding (x-α)(x-β). A classic trap is swapping the sign in the sum or using –c/a for the product.

Sum of roots = -b/a, product = c/a

Memory trick: Sum = -b/a, Prod = c/a.

D = b^2 - 4ac

“When cutting a square cake, the side length is (−b±√D)/(2a); you get a whole-number slice only if √D is an integer and the division by 2a leaves no remainder.”

For integer coefficients, rational roots require D to be a perfect square (√D rational). Integer (integral) roots need, in addition, that 2a divides (−b±√D) so each root is an integer. Trap: mixing the divisibility condition with rational-root criteria.

D = b² - 4ac

Memory trick: Rational: D perfect square. Integer: D perfect square + 2a | (−b±√D).

Vertex at x = -b/2a

“The vertex x=-b/2a is like the balance point of a seesaw where the parabola pivots.”

x-coordinate of the parabola’s extreme is –b/2a; y-value is f(-b/2a). The trap: using –b/2 without dividing by a when the quadratic isn’t monic leads to wrong vertex.

Vertex at x = -b/2a

Memory trick: Balance = –b over 2a.

For roots alpha,beta: (x-alpha)(x-beta)=0

“If a monic quadratic were a lock, each root is a key that opens a linear factor (x-key).”

For a monic quadratic, (x-α)(x-β)=0 gives its roots. When the leading coefficient isn’t 1, the correct form is a(x-α)(x-β). Forgetting the ‘a’ gives wrong constant term.

For roots alpha,beta: (x-alpha)(x-beta)=0

Memory trick: Monic → no extra ‘a’ factor.

|alpha-beta| = sqrt(D)/|a|

“Two tuning-fork frequencies differ by √|D|/|a|, just like the beat frequency from a quadratic’s roots.”

For ax²+bx+c, the root difference α−β equals √D/a. Its magnitude is |α−β|=√|D|/|a|, valid for real or complex conjugate roots. Forgetting the absolute on D gives an imaginary result when D<0, a frequent mistake in JEE problems.

|alpha-beta| = sqrt(D)/|a|

Memory trick: Δroot = √|D| ÷ |a|

real roots vs complex roots

“A parabola crossing the x-axis is like a boat touching shore (real roots); missing it entirely means it sails below (complex).”

Real roots when D≥0 (touch or cross), complex conjugates when D<0. D=0 gives a repeated root – the parabola just kisses the axis. Mistaking D=0 for “no root” loses points.

Memory trick: D>0 cross, D=0 kiss, D<0 miss.

root vs factor

“A root is the exact address where a polynomial vanishes; the factor (x-address) is the building block that contains it.”

Root makes P(x)=0; factor theorem says (x-root) divides P(x). Repeated roots mean repeated linear factors. Assuming each root gives a distinct factor ignores multiplicity.

Memory trick: Root ↔ (x-root) factor; multiplicity repeats.

maximum vs minimum of a parabola

“Rolling a ball into a bowl, it stops at the bowl’s deepest point—the vertex of an upward-opening parabola.”

For ax²+bx+c the vertex y-value is –Δ⁄4a. If a>0 the parabola opens upward, so the vertex is a minimum; if a<0 it opens downward, so it is a maximum. Forgetting the sign of a or applying the formula when a=0 (linear) leads to a wrong conclusion.

Memory trick: a>0 ⇒ min, a<0 ⇒ max; think ‘a’ as the arch direction.

Memory trick: D<0 ⇒ 2 complex conjugates, not 1 real + 1 complex

Argand plane

“Like plotting temperature (x) and humidity (y) on a weather map; each point is a complex number.”

A complex number a+ib is plotted as the point (a,b) on the Argand plane; the distance from the origin O(0,0) is its modulus |z|=√(a²+b²). The plane follows ordinary Euclidean geometry. Mistake: treating i as a mere symbol and ignoring the perpendicular imaginary axis.

Memory trick: a+ib → (a,b) coordinates.

Argand plane — common mistake

“Like assuming a 2-D map is just a line; you miss the north-south axis.”

Many students treat i only as an algebraic symbol, forgetting it represents a unit vector 90° to the real axis. This leads to wrong loci and argument results. Always keep the geometric picture: i rotates a vector by a right angle.

Memory trick: i = 90° turn on the plane.

Modulus and argument

“Like a compass needle, the modulus is the arrow’s length and the argument is its direction.”

In polar form z = r(cosθ + i sinθ), r gives size and θ gives direction. When multiplying, r₁r₂ is the new size and θ₁+θ₂ the new direction. A trap is mixing degrees with radians or forgetting to keep the resultant angle within [0,2π].

Memory trick: Modulus ×, Argument +

Modulus and argument — common mistake

“When two gear wheels mesh, the overall gear ratio is the product of their individual ratios, just like multiplying moduli of complex numbers.”

For z₁·z₂, |z₁z₂| = |z₁|·|z₂| and arg(z₁z₂)=arg(z₁)+arg(z₂). The common slip is to add the moduli. Remember that division uses |z₁|/|z₂| and subtracts arguments; also keep angles in the same unit (rad or deg).

Memory trick: Product: M×, A+; Quotient: M÷, A−

De Moivre's theorem

“Rotating a turntable n times is exactly (cisθ)ⁿ = cis(nθ).”

De Moivre’s theorem states (cisθ)ⁿ = cis(nθ), so raising to a power multiplies the angle while the radius stays rⁿ. It also yields the n-th roots by dividing the angle. Beware: the formula holds for integer n; for non-integers you must use the principal value.

Memory trick: Power → n×θ