Nature of roots

“A parabolic bridge arch meets the river surface: two contact points if D>0, just touches if D=0, floats above if D<0.”

The discriminant D = b²-4ac decides root nature: D>0 gives two distinct real roots, D=0 gives one repeated real root, D<0 gives two complex conjugate roots. This follows directly from solving ax²+bx+c=0. In JEE, compute D first; a zero discriminant still yields a real root, not “no solution”.

Memory trick: D>0 → 2 real; D=0 → 1 real; D<0 → 0 real (complex)

Nature of roots — common mistake

“A doctor checks temperature to know fever, not run full tests; similarly you check D to know root nature.”

Only the sign of D matters for the nature of roots; you need not find the roots themselves. Many waste time solving the quadratic when the question asks merely whether the roots are real, equal or complex. Check D and stop.

Memory trick: Ask D, not x.

x = [−b ± √(b²−4ac)] / 2a

“Your GPS computes the quickest turn by solving a quadratic with x = [–b ± √(b²−4ac)]/2a, just like the formula.”

The quadratic formula x = [–b ± √(b²−4ac)]/2a gives the two solutions of any quadratic ax²+bx+c=0 (a≠0). It is derived from completing the square and works for any coefficients. A frequent error is to miss the ‘2a’ denominator or change the sign of b.

x = [−b ± √(b²−4ac)] / 2a

Memory trick: Minus b ± √(b²−4ac) over 2a.

sum = −b/a, product = c/a

“When two friends split a pizza, the total slices they claim follow sum = –b/a and each gets c/a of the whole.”

Vieta’s relations state α+β = –b/a and α·β = c/a for roots α, β of ax²+bx+c=0. They let you construct a quadratic when the sum or product of roots is known, and vice-versa. Watch out for sign mistakes when b or c are negative.

sum = −b/a, product = c/a

Memory trick: Sum –b/a, Prod c/a – signs matter.

real roots vs complex roots

“A ball thrown on a flat floor lands (real roots) only if D≥0; if D<0 it never touches the ground, giving complex roots.”

If D≥0 the parabola y = ax²+bx+c meets the x-axis, giving real roots (touching when D=0). If D<0 the curve stays above or below the axis, so the solutions are complex conjugates. Remember D=0 still yields a single real root, not two.

Memory trick: D≥0 hits axis, D<0 flies.

roots vs factors

“Factoring x²-5x+6 as (x-2)(x-3) shows 2 and 3 are the x-values that make the quadratic zero.”

Roots are the x-values that satisfy the equation =0. Each linear factor (x-root) corresponds to one root, so a quadratic = a(x-α)(x-β). Beware: include the leading coefficient a when writing factors.

Memory trick: Root → zero; Factor = (x-root)

Myth: D<0 means no solution

“The equation x²+1=0 never touches the x-axis, yet its solutions are i and –i.”

A negative discriminant (D<0) gives two complex conjugate roots, not “no roots”. The quadratic always has two roots in the complex plane. The trap is to equate “no real root” with “no root”.

Memory trick: D<0 → imaginary pair

The imaginary unit

“Multiplying a complex number by i turns it 90° counter-clockwise on the Argand diagram, like rotating a vector on a compass.”

The imaginary unit i is defined by i² = –1, giving a way to take square roots of negative real numbers. In the complex plane, multiplying by i corresponds to a 90° counter-clockwise rotation, so applying it twice flips the direction (i² = –1). On exams never use √a·√b = √(ab) when a or b is negative; that rule holds only for non-negative reals.

Memory trick: i = 90° CCW turn; i² = flip (-1)

The imaginary unit — common mistake

“Treating √(–1)·√(–1) as √1 is like assuming two left-hand turns make a right turn; they actually give a left turn.”

The identity √a·√b = √(ab) holds only for a,b ≥ 0. For a = b = –1, √(–1)·√(–1)=i·i = –1, not +1. Hence √(–1)·√(–1)= –1, confirming i² = –1.

Memory trick: √(–1)·√(–1)= –1 because i²=–1

Modulus and conjugate

“|z| is the straight-line distance from the origin to the point (a,b) on the Argand plane.”

For z = a+bi, modulus |z| = √(a²+b²) is its distance from the origin. Its conjugate ̅z = a–bi reflects across the real axis. Multiplying z by ̅z gives |z|², a real number, useful for rationalising denominators.

Memory trick: z·̅z = |z|²

Modulus and conjugate — common mistake

“Skipping the conjugate when dividing is like forgetting to cancel a common factor; the result stays complex.”

A frequent slip is dividing by a complex number without using its conjugate, leaving an imaginary denominator. Use z/ w = (z·̅w)/|w|² to make the denominator real. Forgetting this step yields an incorrect answer.

Memory trick: Use conjugate → real denominator

Argand plane

“When you plot the voltage phasor of a household AC outlet, its real and imaginary parts become x-y coordinates on the Argand plane.”

A complex number a+bi is represented by the point (a,b) on the Argand plane: real part on the horizontal axis, imaginary coefficient on the vertical. Swapping the axes leaves the distance √(a²+b²) unchanged but gives a different angle. Exam trap: mixing axes changes the argument (quadrant) while the modulus stays the same.

Memory trick: Re → right, Im → up

Argand plane — common mistake

“Imagine drawing a wind vector on a weather map; the east-west component is the x-axis and the north-south component is the y-axis – just like real and imaginary parts.”

The correct plot is (a,b): a on the x-axis, b on the y-axis. Plotting the imaginary part on the x-axis rotates the point, altering the argument and possibly the quadrant, though the modulus √(a²+b²) remains unchanged. Trap: a wrong axis placement gives a wrong angle, not a wrong length.

Memory trick: i goes vertical

|z| = √(a²+b²)

“The modulus of a+bi is the diagonal of a right-triangle with sides a and b, just like the hypotenuse in Pythagoras.”

|z| = √(a²+b²) gives the distance from the origin to (a,b). It follows directly from the Pythagorean theorem. Remember that signs of a or b do not affect the modulus; only their squares matter.

|z| = √(a²+b²)

Memory trick: Pythagoras for complex

z·z̄ = |z|²

“To simplify (3+4i)/(2−i), you multiply top and bottom by 2+i; the denominator becomes (2)²+(1)² = |2−i|², a pure real number.”

For any z = a+bi, z·z̄ = (a+bi)(a−bi)=a²+b² = |z|², a real quantity. This identity is used to rationalise denominators and to find the modulus. Trap: forgetting the minus sign in the conjugate gives (a+bi)², which is not real.

z·z̄ = |z|²

Memory trick: Conjugate flips sign → a²+b²

real part vs imaginary part

“Like a GPS coordinate, a+bi’s ‘a’ tells how far east-west you are, while ‘b’ tells the north-south height on the Argand map.”

In a+bi, a is the real part plotted on the horizontal axis and b is the imaginary part plotted on the vertical axis. The real part is the x-coordinate, the imaginary part the y-coordinate. Beware: the imaginary part is b, not bi; i belongs to the axis, not the value.

Memory trick: Real = R → Right-left axis; Imag = I → Up-ward axis.

modulus vs argument

“A boat’s rope length from the dock is the modulus, and the angle it makes with the dock’s line is the argument of the complex number.”

Modulus |z| = √(a²+b²) is the distance from the origin. Argument θ = atan2(b,a), which automatically accounts for the correct quadrant; using tan⁻¹(b/a) alone can give the wrong principal value when a<0 or a=0. Trap: ignore quadrant → wrong θ.

Memory trick: θ = atan2(b,a) – check quadrant

Myth: √(−1)·√(−1) = 1

“When you flip a coin twice, you still get heads or tails—not a neutral side; likewise √(-1)·√(-1) stays –1, not 1.”

The rule √a·√b = √(ab) holds only for non-negative real a and b. For negative radicands we define √(-1)=i (principal root), so i·i = i² = -1. Using the product rule with negatives gives the wrong answer, a frequent exam trap.

Memory trick: √a·√b = √(ab) only if a,b ≥0

Arithmetic progression — common mistake

“If you start at 5 m and move 3 m each minute, after 4 minutes you are at 5+3×(4-1)=14 m, not 5+3×4.”

In an AP the nth term is a+(n-1)d because the first term is counted as step 1. Using n instead of n-1 overestimates the term. Ensure you count the number of differences, not the number of terms.

Memory trick: n-1 = steps after the first term.

Geometric progression — common mistake

“Doubling a bacteria culture each hour multiplies the count by 2 every step, never adds 2.”

In a GP each term is obtained by multiplying the previous term by the common ratio r, so Tₙ = a·r^(n-1). Adding r each step is wrong. The trap is to treat a linear progression as geometric.

Memory trick: GP = start × (ratio)^(steps).

Infinite geometric series

“When you keep adding half the previous amount of water to a tank, the total water approaches a/(1-r).”

For a GP a, ar, ar²… the sum S = a/(1−r) only if |r|<1, because then the terms shrink to zero and the series converges. If |r|≥1 the terms do not vanish, the series diverges and the formula fails.

Memory trick: Finite sum ⇔ |r|<1 (|r| less than one, series settles)

Infinite geometric series — common mistake

“An echo that loses 30 % of its loudness each bounce has r=0.7; if r≥1 the total sound energy would diverge.”

A common error is using the infinite-GP sum formula when |r|≥1. The series a/(1-r) is valid only for |r|<1; otherwise the terms do not diminish and the sum is infinite.

Memory trick: Never apply a/(1-r) when |r|≥1.

aₙ = a + (n−1)d

“Like climbing stairs where each step is exactly 3 cm higher, the nth step height follows aₙ = a + (n−1)d.”

The nth term of an arithmetic progression (AP) is aₙ = a + (n−1)d, where a is the first term and d the constant difference. It applies only when consecutive terms differ by the same amount. Using it on a geometric pattern gives a wrong answer.

aₙ = a + (n−1)d

Memory trick: AP = Add Progression → add d each step

aₙ = a·r^(n−1)

“A bank account that doubles every year follows aₙ = a·r^(n−1).”

For a geometric progression (GP) the nth term is aₙ = a·r^(n−1), with a the first term and r the common ratio. It works only when each term is obtained by multiplying the previous one by the same r. Mixing r with d is a common error.

aₙ = a·r^(n−1)

Memory trick: GP = Multiply Progression → multiply by r each step