Polynomial, Quadratic and Root Relations Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Solve 11+ Polynomial, Quadratic and Root Relations previous year questions for CAT with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Polynomial, Quadratic and Root Relations

Chapter journey

Polynomial, Quadratic and Root Relations

🌱 Topic 1: Quadratic Roots and Vieta Relations

You learn to convert roots into two numbers: sum and product. This is the selected topic.

CAT PYQ count in this topic: 5 | Importance: moderate but core

🔍 Topic 2: Discriminant, Integer Roots and Root Conditions

You decide what type of roots exist using conditions like real roots, no real roots, equal roots, and integer roots.

CAT PYQ count in this topic: 4

🧩 Topic 3: Polynomial Roots and Conjugate Surds

You extend root logic to polynomial expressions and special root pairs such as conjugate surds.

CAT PYQ count in this topic: 5

By the end of this chapter, the goal is simple: do not fear roots. You should know when to solve, when to use sum-product, and when to use root conditions.

Topic Hero: Roots Without Solving

Selected Topic

Quadratic Roots and Vieta Relations

CAT often hides the roots. Your job is to extract what matters without wasting time.

Main conversion

If roots are

α, β

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

then

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

See roots?

Think sum and product.

See reciprocals?

Divide by product.

See common root?

Let it be $r$ .

Polynomial, Quadratic and Root Relations: Solved Questions with Step-by-Step Explanations (5 Problems)

Question 1 · Quantitative Ability MCQ

Let a, b, c be non-zero real numbers such that $b^{2} < 4 a c$ , and $f (x) = a x^{2} + b x + c$ . If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be

A.
the set of all positive integers
B.
the set of all integers
C.
either the empty set or the set of all integers
D.
the empty set

Correct Answer:

Step-by-Step Solution

Key idea: This is a root condition question based on the discriminant. Recognisable because it asks about the set of integers where a quadratic is negative, given a condition on its coefficients.

Step 1: Interpret the discriminant condition.

The given quadratic is $f (x) = a x^{2} + b x + c$ .

We are given $b^{2} < 4 a c$ , which means the discriminant $D = b^{2} - 4 a c < 0$ .

Since $D < 0$ , the quadratic equation has no real roots. This means the graph of $f (x)$ never crosses the x-axis. Thus, $f (x)$ maintains the same sign for all real numbers $x$ .

Step 2: Analyze the two possible cases for $a$ .

The sign of $f (x)$ for all real $x$ is entirely determined by the leading coefficient $a$ .

Case 1: $a > 0$ .

If $a$ is positive, then $f (x) > 0$ for all real $x$ . In this case, there is no integer $m$ such that $f (m) < 0$ . So the set $S$ is the empty set.

Case 2: $a < 0$ .

If $a$ is negative, then $f (x) < 0$ for all real $x$ . In this case, every integer $m$ satisfies $f (m) < 0$ . So the set $S$ is the set of all integers.

Step 3: Conclusion.

Depending on the unknown sign of $a$ , $S$ must necessarily be either the empty set or the set of all integers.

Answer: either the empty set or the set of all integers

Question 2 · Quantitative Ability NAT

Let $α$ and $β$ be the two distinct roots of the equation $2 x^{2} - 6 x + k = 0$ , such that $(α + β)$ and $α β$ are the distinct roots of the equation $x^{2} + p x + p = 0$ . Then, the value of 8(k - p) is

Correct Answer:

Step-by-Step Solution

Key idea: this is a "roots become roots" Vieta question. The first quadratic gives a sum and a product, and those two quantities are themselves the roots of the second quadratic. So we should use Vieta formulas twice, not solve for the individual roots.

Step 1: For $2 x^{2} - 6 x + k = 0$ , let

$S = α + β, P = α β .$

By Vieta formulas,

$S = - \frac{- 6}{2} = 3, P = \frac{k}{2} .$

Step 2: The second equation is $x^{2} + p x + p = 0$ . Its roots are given as $S$ and $P$ . For this equation, again by Vieta,

$sum of roots = - p, product of roots = p .$

Step 3: Therefore,

$S + P = - p$

and

$S P = p .$

Substitute $S = 3$ and $P = \frac{k}{2}$ :

$3 + \frac{k}{2} = - p$

and

$3 \cdot \frac{k}{2} = p .$

$p = \frac{3 k}{2} .$

Step 4: Put $p = \frac{3 k}{2}$ into the sum equation:

$3 + \frac{k}{2} = - \frac{3 k}{2} .$

Multiply by 2:

$6 + k = - 3 k .$

Hence

$4 k = - 6$

and

$k = - \frac{3}{2} .$

Step 5: Now find $p$ :

$p = \frac{3 k}{2} = \frac{3}{2} \cdot (- \frac{3}{2}) = - \frac{9}{4} .$

Step 6: Compute $k - p$ :

$k - p = - \frac{3}{2} - (- \frac{9}{4})$

$= - \frac{6}{4} + \frac{9}{4} = \frac{3}{4} .$

Therefore,

$8 (k - p) = 8 \cdot \frac{3}{4} = 6.$

Answer: 6

Question 3 · Quantitative Ability MCQ

Suppose k is any integer such that the equation $2 x^{2} + k x + 5 = 0$ has no real roots and the equation $x^{2} + (k - 5) x + 1 = 0$ has two distinct real roots for x. Then, the number of possible values of k is

A.
9
B.
7
C.
8
D.
13

Correct Answer:

Step-by-Step Solution

Key idea: this is a discriminant-condition question. The phrases "no real roots" and "two distinct real roots" must be translated into inequalities involving the discriminant $D = b^{2} - 4 a c$ .

Step 1: For the first equation,

$2 x^{2} + k x + 5 = 0,$

the coefficients are $a = 2$ , $b = k$ , $c = 5$ . Its discriminant is

$D_{1} = k^{2} - 4 (2) (5) = k^{2} - 40.$

"No real roots" means

$D_{1} < 0.$

$k^{2} - 40 < 0$

$k^{2} < 40.$

Since $k$ is an integer,

$- 6 \leq k \leq 6.$

Step 2: For the second equation,

$x^{2} + (k - 5) x + 1 = 0,$

the coefficients are $a = 1$ , $b = k - 5$ , $c = 1$ . Its discriminant is

$D_{2} = (k - 5)^{2} - 4 (1) (1) = (k - 5)^{2} - 4.$

"Two distinct real roots" means

$D_{2} > 0.$

$(k - 5)^{2} - 4 > 0$

$(k - 5)^{2} > 4.$

This gives

$∣ k - 5∣ > 2.$

Hence

$k - 5 > 2 or k - 5 < - 2.$

Therefore,

$k > 7 or k < 3.$

Step 3: Combine both conditions. From the first condition,

$k \in {- 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6} .$

From the second condition,

$k < 3 or k > 7.$

Intersecting with $- 6 \leq k \leq 6$ , the possible integers are

$- 6, - 5, - 4, - 3, - 2, - 1, 0, 1, 2.$

Step 4: Count them:

$9 values .$

Answer: A

Question 4 · Quantitative Ability MCQ

If $(3 + 22)$ is a root of the equation $a x^{2} + b x + c = 0$ and $(4 + 23)$ is a root of the equation $a y^{2} + m y + n = 0$ where a, b, c, m and n are integers, then the value of $(\frac{b}{m} + \frac{c - 2 b}{n})$ is

A.
0
B.
1
C.
3
D.
4

Correct Answer:

Step-by-Step Solution

This is a conjugate surd roots with shared leading coefficient question, recognisable because irrational roots with integer coefficients force conjugate pairs.

Step 1: Apply the conjugate root theorem to the first equation.

Since $3 + 22$ is a root of $a x^{2} + b x + c = 0$ with integer coefficients, $3 - 22$ is also a root.

Sum of roots: $(3 + 22) + (3 - 22) = 6 = - b / a$

Product: $(3 + 22) (3 - 22) = 9 - 8 = 1 = c / a$

So: $b = - 6 a$ and $c = a$ .

Step 2: Apply the conjugate root theorem to the second equation.

Since $4 + 23$ is a root of $a y^{2} + m y + n = 0$ with integer coefficients, $4 - 23$ is also a root.

Sum: $(4 + 23) + (4 - 23) = 8 = - m / a$

Product: $(4 + 23) (4 - 23) = 16 - 12 = 4 = n / a$

So: $m = - 8 a$ and $n = 4 a$ .

Step 3: Compute the required expression.

$\frac{b}{m} = \frac{- 6 a}{- 8 a} = \frac{6}{8} = \frac{3}{4}$

$\frac{c - 2 b}{n} = \frac{a - 2 ( - 6 a )}{4 a} = \frac{a + 12 a}{4 a} = \frac{13 a}{4 a} = \frac{13}{4}$

Step 4: Add.

$\frac{b}{m} + \frac{c - 2 b}{n} = \frac{3}{4} + \frac{13}{4} = \frac{16}{4} = 4$

Answer: 4 (Option D)

Question 5 · Quantitative Ability NAT

A quadratic equation $x^{2} + b x + c = 0$ has two real roots. If the difference between the reciprocals of the roots is $\frac{1}{3}$ , and the sum of the reciprocals of the squares of the roots is $\frac{5}{9}$ , then the largest possible value of $(b + c)$ is

Correct Answer:

Step-by-Step Solution

This is a Vieta's relations with reciprocal roots question, recognisable because the conditions involve reciprocals and reciprocal squares of roots.

Step 1: Set up Vieta's relations.

Let roots be $α$ and $β$ of $x^{2} + b x + c = 0$ .

By Vieta's: $α + β = - b$ and $α β = c$ .

Step 2: Translate the first condition.

Difference of reciprocals: $\frac{1}{α} - \frac{1}{β} = \frac{1}{3}$

$\frac{∣ β - α ∣}{∣ α β ∣} = \frac{1}{3} ⟹ \frac{∣ α - β ∣}{∣ c ∣} = \frac{1}{3}$

Squaring: $\frac{( α - β ) ^{2}}{c ^{2}} = \frac{1}{9}$

Since $(α - β)^{2} = (α + β)^{2} - 4 α β = b^{2} - 4 c$ :

$b^{2} - 4 c = \frac{c ^{2}}{9} \dots (i)$

Step 3: Translate the second condition.

Sum of reciprocal squares: $\frac{1}{α ^{2}} + \frac{1}{β ^{2}} = \frac{5}{9}$

$\frac{α ^{2} + β ^{2}}{( α β ) ^{2}} = \frac{5}{9} ⟹ \frac{( α + β ) ^{2} - 2 α β}{c ^{2}} = \frac{5}{9}$

$\frac{b ^{2} - 2 c}{c ^{2}} = \frac{5}{9} ⟹ 9 b^{2} - 18 c = 5 c^{2} \dots (ii)$

Step 4: Solve the system.

From (i): $b^{2} = \frac{c ^{2}}{9} + 4 c$ . Substitute into (ii):

$9 (\frac{c ^{2}}{9} + 4 c) - 18 c = 5 c^{2}$

$c^{2} + 36 c - 18 c = 5 c^{2}$

$18 c = 4 c^{2} ⟹ 2 c (2 c - 9) = 0$

Since $c \neq = 0$ (reciprocals exist), $c = \frac{9}{2}$ .

Step 5: Find b.

$b^{2} = \frac{( 9/2 ) ^{2}}{9} + 4 \cdot \frac{9}{2} = \frac{81}{36} + 18 = \frac{9}{4} + 18 = \frac{81}{4}$

$b = \pm \frac{9}{2}$

Step 6: Maximise b + c.

$b + c = \frac{9}{2} + \frac{9}{2} = 9$ (taking $b = + 9/2$ )

or $b + c = - \frac{9}{2} + \frac{9}{2} = 0$ (taking $b = - 9/2$ )

Answer: 9

Polynomial, Quadratic and Root Relations Previous Year Questions (PYQs) for CAT: 11+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Polynomial, Quadratic and Root Relations

Polynomial, Quadratic and Root Relations

Topic Hero: Roots Without Solving

Quadratic Roots and Vieta Relations

Polynomial, Quadratic and Root Relations: Solved Questions with Step-by-Step Explanations (5 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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