MathematicsClass 11ICSE

Why Every Quadratic Equation Has Exactly Two Roots

Use the fundamental theorem of algebra and the quadratic formula, solve equations that reduce to quadratics by substitution, handle equations with a common root, and form new equations from given or transformed roots.

What makes quadratic equations so useful?

Heights of thrown balls, areas of plots and profit curves all lead to equations with an

x^{2}

term. Class 11 goes beyond factorising: it asks how many roots such an equation must have, how disguised equations reduce to quadratics, when two equations share a root, and how to build an equation from known roots.

This part covers the fundamental theorem and quadratic formula, reducible equations, common roots, and forming equations from given roots.

What does the fundamental theorem of algebra say, and how do you use the quadratic formula?

**The fundamental theorem of algebra states that every polynomial equation of degree n has exactly n roots, real or complex, counted with repetition — so

a x^{2} + b x + c = 0

always has two roots, given by

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

.

Where the formula comes from.** Dividing by a and completing the square gives

(x + \frac{b}{2 a})^{2} = \frac{b ^{2} - 4 a c}{4 a ^{2}}

and taking square roots of both sides gives the formula.

Worked example. Solve

2 x^{2} - 7 x + 3 = 0

. Here

a = 2

b = - 7

c = 3

, so

b^{2} - 4 a c = 49 - 24 = 25

and

x = \frac{7 \pm 5}{4} = 3 or \frac{1}{2}

Worked example 2. For

x^{2} + 2 x + 5 = 0

b^{2} - 4 a c = 4 - 20 = - 16

, so

x = \frac{- 2 \pm 4 i}{2} = - 1 \pm 2 i

— two complex roots, just as the theorem promises.

An everyday example. A garden plot 3 m longer than it is wide with an area of 40 square metres gives

x^{2} + 3 x - 40 = 0

, so the width is 5 m; the other root,

- 8

, is rejected because a length cannot be negative.

The substance. A repeated root still counts twice —

x^{2} - 6 x + 9 = 0

has the double root 3.

How do you solve equations that reduce to quadratic form by substitution?

**An equation reduces to a quadratic when a repeated expression — such as

x^{2}

x

2^{x}

— can be replaced by a single new variable; the quadratic is then solved and the substitution reversed.

Common substitutions:**

-

x^{4} - 13 x^{2} + 36 = 0

— put

y = x^{2}

x - 5 x + 6 = 0

— put

y = x

4^{x} - 6 \times 2^{x} + 8 = 0

— put

y = 2^{x}

Worked example. Solve

x^{4} - 13 x^{2} + 36 = 0

. With

y = x^{2}

y^{2} - 13 y + 36 = 0

, so

(y - 4) (y - 9) = 0

and

y = 4

9

. Then

x = \pm 2

\pm 3

— four roots.

An everyday example. Two pipes fill a village water tank; one takes 5 hours longer than the other, and together they take 6 hours. Then

\frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6}

clears to

x^{2} - 7 x - 30 = 0

, so the faster pipe takes 10 hours.

The substance. Reject impossible values of the new variable —

2^{x}

and

x

can never be negative, so a negative y gives no root.

How do you solve problems on quadratic equations with a common root?

**If

a_{1} x^{2} + b_{1} x + c_{1} = 0

and

a_{2} x^{2} + b_{2} x + c_{2} = 0

share a root

α

, treating

α^{2}

and

α

as unknowns and cross-multiplying gives

α = \frac{c _{1} a _{2} - c _{2} a _{1}}{a _{1} b _{2} - a _{2} b _{1}}

and the condition

(c_{1} a_{2} - c_{2} a_{1})^{2} = (b_{1} c_{2} - b_{2} c_{1}) (a_{1} b_{2} - a_{2} b_{1})

; if both roots are common, the coefficients are proportional.

One common root.**

\frac{α ^{2}}{b _{1} c _{2} - b _{2} c _{1}} = \frac{α}{c _{1} a _{2} - c _{2} a _{1}} = \frac{1}{a _{1} b _{2} - a _{2} b _{1}}

Both roots common.

\frac{a _{1}}{a _{2}} = \frac{b _{1}}{b _{2}} = \frac{c _{1}}{c _{2}}

.

Worked example. Find k if

x^{2} - 3 x + 2 = 0

and

x^{2} + k x - 6 = 0

have a common root. The first equation has roots 1 and 2.

- If 1 is common:

1 + k - 6 = 0

, so

k = 5

- If 2 is common:

4 + 2 k - 6 = 0

, so

k = 1

Check

k = 1

with the general formula:

α = \frac{2 - ( - 6 )}{1 - ( - 3 )} = \frac{8}{4} = 2

, the shared root.

An everyday example. Two rectangular plots of different shapes that must share one possible width — each plot's area condition is a quadratic, and the shared width is their common root.

The substance. Try the easy route first — if one equation factorises, test each of its roots in the other before using the general condition.

How do you form a quadratic equation from given roots, including transformed roots such as cubes?

**A quadratic with roots

α

and

β

x^{2} - (α + β) x + α β = 0

, and for transformed roots such as

α^{3}

and

β^{3}

, their sum and product are written in terms of

α + β = - \frac{b}{a}

and

α β = \frac{c}{a}

without solving the original equation.

Useful identities:**

-

α^{2} + β^{2} = (α + β)^{2} - 2 α β

α^{3} + β^{3} = (α + β)^{3} - 3 α β (α + β)

\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β}

Worked example. If

α

and

β

are the roots of

x^{2} - 3 x + 1 = 0

, form the equation whose roots are

α^{3}

and

β^{3}

.

-

α + β = 3

and

α β = 1

α^{3} + β^{3} = 27 - 3 \times 1 \times 3 = 18

α^{3} β^{3} = 1^{3} = 1

x^{2} - 18 x + 1 = 0

An everyday example. A question-setter who wants the tidy answers 4 and 7 simply writes

x^{2} - 11 x + 28 = 0

, working backwards from the roots.

The substance. Complex roots of a real quadratic come in conjugate pairs — so the roots

2 + i

and

2 - i

give the real equation

x^{2} - 4 x + 5 = 0

Exam tip

What earns full marks on solving and forming quadratic equations?

**Write a, b, c and the value of

b^{2} - 4 a c

on separate lines before substituting into the formula — sign slips in b are the commonest loss.**

-

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

- After a substitution, reverse every value and reject impossible ones
- Common root: test the roots of whichever equation factorises
- New equation:

x^{2} - (sum) x + product = 0

The trap. Writing the equation with roots

α

and

β

x^{2} + (α + β) x + α β = 0

. The sum of the roots enters with a minus sign.

Did you know

Why is there no general formula for equations of degree five?

Quadratic equations have a familiar formula, and cubic and quartic equations also have formulae built from square roots and cube roots — though they are long and rarely used by hand.

For general equations of degree five and above, no such formula can exist. Some of these equations have roots that cannot be written using arithmetic operations and roots of numbers, however they are combined.

The fundamental theorem still guarantees five roots; they simply cannot always be captured in a formula, so computers find them numerically.

Exam relevance

How are quadratic equations tested in JEE Main?

Quadratic Equations is a recurring JEE Main chapter, and relations between roots and coefficients return in complex numbers, sequences and coordinate geometry.

What gets asked. Symmetric expressions in the roots such as

α^{3} + β^{3}

, equations with transformed roots, common-root conditions, and equations reducible to quadratics through exponential or modulus substitutions.

Question types. Multiple-choice and numerical-value questions; JEE Advanced often adds a parameter that must be found.

The trap that costs marks. **Keeping a negative value of

2^{x}

∣ x ∣

** after substituting.

Key takeaways

What must you be able to do from this part?

- Fundamental theorem: a quadratic has exactly two roots, real or complex;

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

- Reducible equations: substitute for the repeated expression, solve, then reverse and check
- Common roots: test factorised roots, or use the cross-multiplication condition
- Forming equations:

x^{2} - (α + β) x + α β = 0

, with transformed roots built from the sum and product

If

α

and

β

are the roots of

x^{2} - 5 x + 6 = 0

, which equation has the roots

α^{2}

and

β^{2}

What makes quadratic equations so useful?

What does the fundamental theorem of algebra say, and how do you use the quadratic formula?

How do you solve equations that reduce to quadratic form by substitution?

How do you solve problems on quadratic equations with a common root?

How do you form a quadratic equation from given roots, including transformed roots such as cubes?

What earns full marks on solving and forming quadratic equations?

Why is there no general formula for equations of degree five?

How are quadratic equations tested in JEE Main?

What must you be able to do from this part?

Related study notes

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