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JEE Mains Maths · Formula sheet

Quadratic Equations formulas

15 formulas and 15 common traps for JEE Mains Maths Quadratic Equations, grouped by subtopic.

Full notes with worked examples

Roots and Coefficients

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Sum, product and difference of roots

Sum, product, difference

α+β=−ba,αβ=ca,(β−α)2=b2−4aca2\alpha+\beta=-\frac{b}{a},\quad \alpha\beta=\frac{c}{a},\quad (\beta-\alpha)^2=\frac{b^2-4ac}{a^2}

Beyond degree two: cubics and polynomial divisors

Vieta for a cubic

α+β+γ=−ba,∑αβ=ca,αβγ=−da\alpha+\beta+\gamma=-\frac{b}{a},\quad \sum\alpha\beta=\frac{c}{a},\quad \alpha\beta\gamma=-\frac{d}{a}

Common traps

The difference needs its square

β−α\beta-\alpha changes sign when the roots swap, so it cannot be written in the sum and product directly. Work with (β−α)2(\beta-\alpha)^2, and take the square root only at the end.

The signs alternate

For a cubic the sum of the roots is −ba-\frac{b}{a}, the pair sum +ca+\frac{c}{a}, and the product −da-\frac{d}{a} — not da\frac{d}{a}. The signs go minus, plus, minus.

Symmetric Functions and Power Sums

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Symmetric expressions from the sum and product

Squares to fourth powers

α4+β4=(α2+β2)2−2(αβ)2\alpha^4+\beta^4=\left(\alpha^2+\beta^2\right)^2-2(\alpha\beta)^2

High powers by a recurrence

Power-sum recurrence

x2=sx−p ⇒ Pn=sPn−1−pPn−2x^2=sx-p\ \Rightarrow\ P_n=sP_{n-1}-pP_{n-2}

Common traps

Check that the roots can exist

Solving for the product often gives two values. For real roots keep only one with s2−4p≥0s^2-4p\ge0; if the numbers must also be positive, the product must be positive too.

Read the sign of the product

For x2−sx+p=0x^2-sx+p=0 the rule is Pn=sPn−1−pPn−2P_n=sP_{n-1}-pP_{n-2}. With x2−2x−3=0x^2-2x-3=0, p=−3p=-3, so Pn=2Pn−1+3Pn−2P_n=2P_{n-1}+3P_{n-2}: the minus sign becomes a plus.

Common Roots and New Equations

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A root shared by two equations

Condition for one common root

(c1a2−c2a1)2=(a1b2−a2b1)(b1c2−b2c1)(c_1a_2-c_2a_1)^2=(a_1b_2-a_2b_1)(b_1c_2-b_2c_1)

Building an equation from its roots

Equation from its roots

x2−(r1+r2)x+r1r2=0x^2-(r_1+r_2)x+r_1r_2=0

Common traps

Proportional only for non-real roots

Two equations with real roots can share one root and not the other. Making the coefficients proportional then gives a wrong answer. Use proportionality only when one equation's roots are non-real.

Divide by the leading coefficient

For ax2+bx+c=0ax^2+bx+c=0 the sum of the roots is −ba-\frac ba, not −b-b. Read the sum and product only after dividing by aa, and clear fractions at the end so the answer matches the options.

Discriminant and Location of Roots

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The sign of the discriminant

Discriminant

D=b2−4ac,x=−b±D2aD=b^2-4ac,\qquad x=\frac{-b\pm\sqrt D}{2a}

Rational and integer roots

Rational roots

b2−4ac∈Z ⇒ x=−b±D2a is rational\sqrt{b^2-4ac}\in\mathbb{Z}\ \Rightarrow\ x=\frac{-b\pm\sqrt{D}}{2a}\ \text{is rational}

Placing the roots on the number line

A number between the roots

α<k<βexactly whena f(k)<0\alpha<k<\beta\quad\text{exactly when}\quad a\,f(k)<0

Common traps

The x² coefficient can vanish

When the leading coefficient holds a parameter, the value that makes it 00 leaves a linear equation, and the discriminant test does not apply. Treat that value on its own — it is often the case the options exclude.

Rational coefficients first

The perfect-square test works only when the coefficients are rational. With 2\sqrt2 in a coefficient, the roots can be irrational even when DD is a perfect square.

The vertex condition is not optional

D≥0D\ge0 and f(k)>0f(k)>0 hold both when the two roots are above kk and when both are below it. Only the vertex −b2a-\frac{b}{2a} tells which side.

Equations with Modulus and Greatest Integer

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Splitting at the critical points

Definition of modulus

∣x−a∣={x−a,x≥aa−x,x<a|x-a|=\begin{cases}x-a,& x\ge a\\ a-x,& x<a\end{cases}

One modulus as the unknown

Equality in the triangle inequality

∣A+B∣=∣A∣+∣B∣exactly whenAB≥0|A+B|=|A|+|B|\quad\text{exactly when}\quad AB\ge0

Greatest integer and fractional part

Greatest integer

[x]=nexactly whenn≤x<n+1[x]=n\quad\text{exactly when}\quad n\le x<n+1

Common traps

A root on the boundary is counted once

A critical point can solve the equation in both neighbouring pieces. Count it once. And reject a root from a piece it does not lie in, even though it solves that piece's quadratic.

Reject negative values of |x|

The quadratic in t=∣x∣t=|x| can have a negative root, and it gives no xx at all. Count roots only from t≥0t\ge0, and remember that t=0t=0 gives one root, not two.

The fractional part is below 1

{x}\{x\} is never 11 or more and never negative. A factor that forces {x}=3\{x\}=3, or any value outside [0,1)[0,1), gives no root.

Equations Reducible to Quadratics

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Exponential equations

Reciprocal pair

t+1t=k ⇒ t=k±k2−42t+\frac1t=k\ \Rightarrow\ t=\frac{k\pm\sqrt{k^2-4}}{2}

Logarithmic equations

Swapped bases

log⁡ba=1log⁡ab\log_b a=\frac{1}{\log_a b}

A repeated expression as the unknown

Reciprocal substitution

x2+1x2=(x+1x)2−2x^2+\frac1{x^2}=\left(x+\frac1x\right)^2-2

Common traps

Every t must be positive

axa^x is never 00 or negative, so such a root in tt gives no xx. The sum of the xx-roots is the log of the product of the valid tt-roots only — not of every root of the polynomial in tt.

Check every base

A root can make a base negative or equal to 1 even when every argument is positive. Test each root in each base and each argument before counting it.

Undo the substitution with its range

Each root in tt must be a value its block can take. x+1xx+\frac1x never lies strictly between −2-2 and 22, and x\sqrt x and x2x^2 are never negative; such roots give no xx.

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