MathematicsClass 10CBSE

You Can Read Both Zeroes of a Quadratic Straight Off Its Coefficients

Find the zeroes of a quadratic by factorisation, verify them with the sum and product relations, build a polynomial from a given sum and product, and use the same relations to find an unknown coefficient without solving anything.

Why can you check your zeroes without substituting them back?

Solve

x^{2} - 3 x - 10 = 0

and you get

x = 5

and

x = - 2

. Now look at what those two answers do together:

- **They add to

3

** — and the coefficient of

x

- 3

- **They multiply to

- 10

** — and the constant term is

- 10

That is not a coincidence about this one polynomial. The sum and the product of the zeroes of any quadratic are fixed by its coefficients, and you can read them off before you have solved anything.

Which gives you two powerful things at once.

- A check. Every pair of zeroes you find can be verified in two lines, without substituting either value back into the polynomial
- A shortcut. A great many questions ask only for the sum, only for the product, or for a coefficient that makes the zeroes satisfy some condition — and none of those require finding the zeroes at all

The chapter also gives the zeroes a picture. **A zero of a polynomial is exactly a point where its graph crosses the

x

-axis, so the number of zeroes can be read off a sketch. A quadratic's graph is a parabola, and it can meet the axis twice, once, or not at all — which is the geometrical meaning of having two, one or no real zeroes.

And the same relations extend upward. A cubic has three zeroes and three relations**, built the same way, which is where the chapter ends.

This page covers the CBSE Class 10 Maths chapter on polynomials: the geometrical meaning of zeroes, the relations between zeroes and coefficients for quadratics and cubics, and constructing a polynomial from given zeroes.

Formula

What are the relations between the zeroes and the coefficients?

**For a quadratic, the sum of the zeroes is minus the coefficient of

x

over the coefficient of

x^{2}

, and the product is the constant term over the coefficient of

x^{2}

.**

For

p (x) = a x^{2} + b x + c

with zeroes

α

and

β

α + β = - \frac{b}{a}, α β = \frac{c}{a}

Worked example 1 — find and verify. Find the zeroes of

x^{2} - 3 x - 10

and verify both relations.

Factorise by splitting the middle term. We need two numbers whose product is

- 10

and whose sum is

- 3

: those are

- 5

and

2

x^{2} - 3 x - 10 = x^{2} - 5 x + 2 x - 10 = x (x - 5) + 2 (x - 5) = (x - 5) (x + 2)

So the zeroes are

α = 5

and

β = - 2

.

Verify the sum. Here

a = 1

b = - 3

c = - 10

α + β = 5 + (- 2) = 3, - \frac{b}{a} = - \frac{- 3}{1} = 3

Verify the product.

α β = 5 \times (- 2) = - 10, \frac{c}{a} = \frac{- 10}{1} = - 10

Both match.

**Worked example 2 — with a leading coefficient that is not

1

.** Find the zeroes of

6 x^{2} - 7 x - 3

and verify the relations.

We need two numbers whose product is

6 \times (- 3) = - 18

and whose sum is

- 7

: those are

- 9

and

2

6 x^{2} - 7 x - 3 = 6 x^{2} - 9 x + 2 x - 3 = 3 x (2 x - 3) + 1 (2 x - 3) = (2 x - 3) (3 x + 1)

Setting each factor to zero:

2 x - 3 = 0 \Rightarrow x = \frac{3}{2}, 3 x + 1 = 0 \Rightarrow x = - \frac{1}{3}

Verify the sum. Here

a = 6

b = - 7

c = - 3

α + β = \frac{3}{2} - \frac{1}{3} = \frac{9 - 2}{6} = \frac{7}{6}, - \frac{b}{a} = - \frac{- 7}{6} = \frac{7}{6}

Verify the product.

α β = \frac{3}{2} \times (- \frac{1}{3}) = - \frac{3}{6} = - \frac{1}{2}, \frac{c}{a} = \frac{- 3}{6} = - \frac{1}{2}

Both match again, and notice that the fractions were the place an error could creep in — which is exactly why the check is worth two lines.

The cubic relations. For

p (x) = a x^{3} + b x^{2} + c x + d

with zeroes

α

β

γ

α + β + γ = - \frac{b}{a}

α β + β γ + γ α = \frac{c}{a}

α β γ = - \frac{d}{a}

Read the pattern and you will not need to memorise them separately. The signs alternate starting with minus, and each relation takes the next coefficient: the sum of the zeroes, then the sum of the products in pairs, then the product of all three.

Worked example 3 — a cubic. For

2 x^{3} - 5 x^{2} - 14 x + 8

, write down the three symmetric quantities without finding a single zero.

Here

a = 2

b = - 5

c = - 14

d = 8

α + β + γ = - \frac{- 5}{2} = \frac{5}{2}

α β + β γ + γ α = \frac{- 14}{2} = - 7

α β γ = - \frac{8}{2} = - 4

Three answers, no solving. That is the point of the relations.

How do you build a quadratic from a given sum and product?

**Write

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

, then clear the fractions if there are any.**

p (x) = k [x^{2} - (α + β) x + α β]

for any non-zero constant

k

. **Taking

k = 1

gives the simplest such polynomial**, and the reason a

k

appears at all is worth stating: multiplying a polynomial by a constant does not change where it is zero, so infinitely many quadratics have the same pair of zeroes.

Notice the minus sign in front of the sum. It comes straight from

α + β = - \frac{b}{a}

, and forgetting it is the commonest error in this topic.

Worked example 1. Find a quadratic polynomial whose zeroes have a sum of

- 3

and a product of

2

p (x) = x^{2} - (- 3) x + 2 = x^{2} + 3 x + 2

Check by factorising:

x^{2} + 3 x + 2 = (x + 1) (x + 2)

, giving zeroes

- 1

and

- 2

. Their sum is

- 3

and their product is

2

. Both requirements met.

Worked example 2 — with fractions. Find a quadratic polynomial whose zeroes have a sum of

\frac{1}{4}

and a product of

- 1

p (x) = x^{2} - \frac{1}{4} x - 1

Multiplying through by

4

to clear the fraction:

p (x) = 4 x^{2} - x - 4

Check with the relations:

a = 4

b = - 1

c = - 4

, so

- \frac{b}{a} = \frac{1}{4} and \frac{c}{a} = \frac{- 4}{4} = - 1

Both are as required. Clearing the denominator was allowed precisely because the

k

is free.

Worked example 3 — from the zeroes themselves. Find a quadratic polynomial with zeroes

2 + 3

and

2 - 3

.

First the sum and the product:

sum = (2 + 3) + (2 - 3) = 4

product = (2 + 3) (2 - 3) = 4 - 3 = 1

p (x) = x^{2} - 4 x + 1

Notice how the surds disappeared. Because the two zeroes are conjugates, the sum and product are both rational — which is why a quadratic with rational coefficients can have irrational zeroes only in such a pair. That observation is examined directly.

Worked example 4 — the cubic version. Find a cubic polynomial whose zeroes have a sum of

2

, a sum of products in pairs of

- 7

, and a product of

- 14

.

The pattern extends with alternating signs:

p (x) = x^{3} - 2 x^{2} - 7 x + 14

The signs, read left to right, are minus, plus, minus applied to the three given quantities — which is exactly the alternation in the cubic relations.

How do you find an unknown coefficient from a condition on the zeroes?

Translate the condition into a statement about the sum or the product, then read that off the coefficients. You almost never need to find the zeroes.

Worked example 1 — both zeroes given. If

2

and

- 3

are the zeroes of

x^{2} + (a + 1) x + b

, find

a

and

b

.

Here the leading coefficient is

1

, so the sum of the zeroes is

- (a + 1)

and the product is

b

.

From the sum:

2 + (- 3) = - 1 ⟹ - (a + 1) = - 1 ⟹ a + 1 = 1 ⟹ a = 0

From the product:

2 \times (- 3) = - 6 ⟹ b = - 6

Check by substituting back. With

a = 0

and

b = - 6

the polynomial is

x^{2} + x - 6 = (x + 3) (x - 2)

, whose zeroes are

- 3

and

2

. Correct.

Worked example 2 — reciprocal zeroes. If the zeroes of

2 x^{2} + 3 x + k

are reciprocals of each other, find

k

.

If the zeroes are

α

and

\frac{1}{α}

, then their **product is

1

.** So

\frac{c}{a} = 1 ⟹ \frac{k}{2} = 1 ⟹ k = 2

Notice that the sum was never needed. "Reciprocal zeroes" is a statement about the product alone, and recognising that is the whole solution.

Worked example 3 — equal and opposite zeroes. If the zeroes of

x^{2} + p x + 9

are equal in magnitude but opposite in sign, find

p

.

If the zeroes are

α

and

- α

, their **sum is

0

.** So

- \frac{b}{a} = 0 ⟹ - p = 0 ⟹ p = 0

Again only one relation was needed.

The translation table worth memorising, because every question of this family uses one row of it:

- "Reciprocal zeroes" — the product is

1

- "Equal and opposite zeroes" — the sum is

0

- "One zero is the negative of the other" — the sum is

0

- "One zero is zero" — the product is

0

, so the constant term is

0

- "The zeroes are equal" — the two zeroes are the same number, so the graph touches the axis at one point
- "One zero is given" — substitute it into the polynomial, which must give

0

Worked example 4 — expressions in the zeroes. If

α

and

β

are the zeroes of

x^{2} - 3 x - 10

, find

\frac{1}{α} + \frac{1}{β}

and

α^{2} + β^{2}

without using the zeroes.

From the earlier work,

α + β = 3

and

α β = - 10

. Then

\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{3}{- 10} = - \frac{3}{10}

α^{2} + β^{2} = (α + β)^{2} - 2 α β = 9 - 2 (- 10) = 9 + 20 = 29

**Check against the actual zeroes

5

and

- 2

**:

\frac{1}{5} - \frac{1}{2} = \frac{2 - 5}{10} = - \frac{3}{10}

, and

25 + 4 = 29

. Both agree.

The identity that unlocks most of these is

α^{2} + β^{2} = (α + β)^{2} - 2 α β

. Anything symmetric in the two zeroes can be rewritten using only the sum and the product, which is why the relations are so much more useful than the zeroes themselves.

What does the graph tell you about the number of zeroes?

**The zeroes of a polynomial are exactly the

x

-coordinates of the points where its graph meets the

x

-axis. So counting intersections counts zeroes, and no algebra is needed to answer such a question from a diagram.

The three degrees you must recognise.

- A linear polynomial

a x + b

has a straight-line graph, which meets the axis at exactly one point, so it has exactly one zero
- A quadratic

a x^{2} + b x + c

has a parabola** for its graph — opening upward if

a > 0

and downward if

a < 0

— and it has at most two zeroes
- A cubic has at most three zeroes

**In general a polynomial of degree

n

has at most

n

zeroes, which is the statement the graphs are illustrating.

The three cases for a parabola, and each is a standard diagram question.

- It cuts the

x

-axis at two distinct points — two distinct real zeroes
- It touches the

x

-axis at exactly one point — two equal zeroes, which is counted as one distinct value
- It does not meet the

x

-axis at all — no real zeroes

How to read a given sketch. Count the intersections with the horizontal axis and nothing else. Turning points, the

y

-intercept and how high the curve goes are all irrelevant to the number of zeroes, and questions are deliberately drawn to tempt you into counting them.

Worked reasoning — sign of the leading coefficient.** A parabola opens upward exactly when

a > 0

. So a downward-opening parabola that lies entirely below the

x

-axis has

a < 0

and no real zeroes, while a downward-opening parabola crossing the axis twice has

a < 0

and two zeroes. The direction of opening and the number of zeroes are separate pieces of information, read from different features of the picture.

Worked example — connecting graph and algebra. The graph of

y = x^{2} + 3 x + 2

crosses the

x

-axis at two points. Where?

Factorising,

x^{2} + 3 x + 2 = (x + 1) (x + 2)

, so the crossings are at

x = - 1

and

x = - 2

. **The graph therefore meets the axis at

(- 1, 0)

and

(- 2, 0)

**, and since

a = 1 > 0

the parabola opens upward, dipping below the axis between those two points.

One caution about a cubic's shape. A cubic graph always crosses the axis at least once, because its two ends go in opposite vertical directions. So a cubic can have one, two or three zeroes but never none — unlike a quadratic, which can miss the axis entirely. That contrast is worth holding on to, because it is the kind of one-line reasoning question this chapter likes.

Exam tip

Which habits keep a polynomials answer clean?

Verify with the two relations every single time. They cost two lines and they catch every sign error and every arithmetic slip in a factorisation.

- **Write down

a

b

and

c

explicitly before using the relations, with their signs. Most errors here are sign errors, not method errors
- Remember the minus** in

α + β = - \frac{b}{a}

, and that there is no minus in

α β = \frac{c}{a}

- When building a polynomial, the sum enters with a minus:

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

- Clear fractions at the end by multiplying through, which is allowed because any constant multiple has the same zeroes
- Translate the condition first — reciprocal means product

1

, equal and opposite means sum

0

— and then use only the relation you need
- **Use

α^{2} + β^{2} = (α + β)^{2} - 2 α β

for symmetric expressions rather than finding the zeroes
- For a cubic, keep the alternating signs: minus, plus, minus
- On a graph question, count only the crossings of the

x

-axis

The misconception to name.** The zeroes of a polynomial are not the same thing as its coefficients, and "the zero of

p (x)

" does not mean the value of

p (0)

. **A zero is a value of

x

that makes

p (x) = 0

**;

p (0)

is the constant term. The two get confused because of the shared word, and a question asking for one while you supply the other loses every mark.

A second trap. Assuming a quadratic always has two real zeroes. It can have none, and the graph that never meets the axis is the picture of that case. Writing "the zeroes are" and then producing two numbers for such a polynomial is a marked error.

Did you know

Why do irrational zeroes of a nice quadratic always come in pairs?

Look again at the quadratic built from

2 + 3

and

2 - 3

. Its sum was

4

and its product was

1

, both perfectly rational, and the polynomial came out as

x^{2} - 4 x + 1

with no surd anywhere in sight.

**Now try to build one with only

2 + 3

as a zero and something rational as the other**, say

5

. The sum is

7 + 3

and the product is

10 + 53

, so the polynomial is

x^{2} - (7 + 3) x + (10 + 53)

The surds refuse to leave. There is no way to clear them by multiplying through, because they appear in one coefficient and not in another.

So the conclusion is forced. If a quadratic has rational coefficients and one irrational zero of the form

p + q

, **its other zero must be

p - q

— otherwise the sum and the product could not both be rational. The irrational zeroes come in conjugate pairs, and this is not a convention but a consequence of the relations you have been using all along.

The same argument works for the number of zeroes being irrational at all. A quadratic with rational coefficients cannot have exactly one irrational zero. It has none, or it has two — and the two are conjugates.

Where you have already seen this.** When you rationalise a denominator by multiplying by the conjugate, the surds vanish for exactly this reason:

(p + q) (p - q) = p^{2} - q

, which is rational. The product of a conjugate pair is always rational, and so is their sum. Rationalising a denominator and building a quadratic from conjugate zeroes are the same computation viewed from two sides.

And it carries forward unchanged. In Class 11 you will meet complex zeroes and find that they, too, come in conjugate pairs whenever the coefficients are real — for precisely the same reason. The sum and the product must be real, and only a conjugate pair can deliver that.

One practical use right now. If a question tells you that

3 - 5

is a zero of a quadratic with rational coefficients, you already know the other zero without any working. Sum

= 6

, product

= 9 - 5 = 4

, so the polynomial is

x^{2} - 6 x + 4

. One given zero, a complete answer — and the relations did all of it.

Exam relevance

How do the zero–coefficient relations matter for JEE?

This is foundation work for Class 11 Quadratic Equations and Complex Numbers, and for the theory-of-equations questions in JEE Main and JEE Advanced.

Where the relations lead. Class 11 keeps

α + β = - \frac{b}{a}

and

α β = \frac{c}{a}

unchanged and adds the general theory: for a polynomial of degree

n

the symmetric functions of the roots are read off the coefficients with alternating signs, which is exactly the cubic pattern you learn here extended. The name given to that collection of statements is the theory of symmetric functions, and JEE Advanced sets questions on it directly.

Where the symmetric-expression trick leads. Finding

α^{2} + β^{2}

α^{3} + β^{3}

\frac{1}{α} + \frac{1}{β}

α - β

from the sum and the product is a standard JEE Main item, and the identities multiply:

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

and

(α - β)^{2} = (α + β)^{2} - 4 α β

. The method you practise here — never find the roots, rewrite in terms of the sum and the product — is the whole technique.

Where the graph reading leads. Class 11 turns the three parabola cases into the discriminant:

b^{2} - 4 a c

positive, zero or negative decides two distinct real roots, equal roots, or no real roots. The pictures you learn here are what the discriminant is describing, and a great many JEE questions are answered from the sign of the discriminant plus the position of the vertex.

Where the conjugate-pair result leads. Class 11 proves that complex roots of a real polynomial occur in conjugate pairs, using the same sum-and-product reasoning. The surd version you meet here is the same theorem over the rationals.

Where building a polynomial leads. Constructing an equation whose roots are a transformation of given roots — reciprocals, squares, each increased by a constant — is a recurring competitive item. It is the construction you learn here applied to a new sum and a new product, both computed from the old ones.

Question types to expect. At this level: zeroes by factorisation, verification of the relations, forming a polynomial, finding an unknown coefficient, and reading a graph. In competitive papers: symmetric expressions, transformed equations, discriminant conditions, and the location of roots relative to a given number.

The single trap that costs marks. Dropping the minus sign from the sum relation. **

α + β = - \frac{b}{a}

, never

\frac{b}{a}

— and because a sign error does not look like an error, the verification step is the only thing that catches it.

A second trap. Forgetting that a polynomial is determined only up to a constant multiple. A question asking for "a" quadratic with a given sum and product has infinitely many answers**, and

4 x^{2} - x - 4

is just as correct as

x^{2} - \frac{1}{4} x - 1

. Marking schemes accept either, but a student who thinks the answer is unique will hesitate over clearing fractions.

Board versus competitive emphasis. The CBSE paper marks the factorisation, the two relations and the verification; a competitive paper marks the symmetric expression or the transformed equation. The transferable habit is refusing to find the roots until you are certain you need them — most questions in this family never require them at all.

Key takeaways

What must you be able to do from polynomials?

Two relations for a quadratic, three for a cubic, and one construction.

- **A zero of

p (x)

** is a value of

x

for which

p (x) = 0

, and it is where the graph meets the

x

-axis
- **A polynomial of degree

n

has at most

n

zeroes — one for a linear, at most two for a quadratic, at most three for a cubic
- A parabola cuts the axis twice, touches it once, or misses it entirely, giving two, equal, or no real zeroes
- A cubic always has at least one zero, since its two ends go in opposite directions
- Quadratic relations**:

α + β = - \frac{b}{a}

and

α β = \frac{c}{a}

— mind the minus in the first and its absence in the second
- Cubic relations:

- \frac{b}{a}

, then

\frac{c}{a}

, then

- \frac{d}{a}

, with the signs alternating
- Verify every factorisation with both relations, as

x^{2} - 3 x - 10

gives

5

and

- 2

with sum

3

and product

- 10

- Fractions are where errors hide —

6 x^{2} - 7 x - 3

gives

\frac{3}{2}

and

- \frac{1}{3}

with sum

\frac{7}{6}

and product

- \frac{1}{2}

- To build a quadratic:

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

, then clear fractions, so sum

\frac{1}{4}

and product

- 1

give

4 x^{2} - x - 4

- Translate conditions: reciprocal zeroes means product

1

, so

2 x^{2} + 3 x + k

needs

k = 2

; equal and opposite means sum

0

- Symmetric expressions come from

α^{2} + β^{2} = (α + β)^{2} - 2 α β

and

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

- Irrational zeroes of a rational quadratic come in conjugate pairs, so one given zero determines the other

The sharpest self-test is one you can set yourself. Pick any two numbers, build the quadratic from their sum and product, then factorise your own polynomial and see whether the zeroes you get back are the two you started with.

Why can you check your zeroes without substituting them back?

What are the relations between the zeroes and the coefficients?

How do you build a quadratic from a given sum and product?

How do you find an unknown coefficient from a condition on the zeroes?

What does the graph tell you about the number of zeroes?

Which habits keep a polynomials answer clean?

Why do irrational zeroes of a nice quadratic always come in pairs?

How do the zero–coefficient relations matter for JEE?

What must you be able to do from polynomials?

Related study notes

Ready to put this into practice?