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Samacheer Kalvi 12th Maths Chapter 3 Exercise Solutions

How to Use This Page

This page is your complete study companion for Chapter 3 — Theory of Equations. Start with the Key Concepts table to refresh definitions and formulas, then work through the MCQs, fill-in-the-blanks, and true-or-false sections for quick practice. Use the match-the-following exercise to strengthen associations. The brief and detailed answer sections mirror the 2-mark and 5-mark questions you will face in the board exam. Finally, review the Common Mistakes and Exam Checklist before every test to avoid losing marks.

Chapter 3 — Exercise Solutions Guide

This chapter is formula-based. First write the relation between roots and coefficients, then substitute carefully.

Key Concepts (Quick Recap)

Term	Definition
Polynomial equation	An equation of the form a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0 = 0
Root (zero)	A value of x that satisfies the polynomial equation
Vieta's formulas (quadratic)	For ax² + bx + c = 0: sum of roots = -b/a, product = c/a
Vieta's formulas (cubic)	For ax³ + bx² + cx + d = 0: sum = -b/a, sum of pairwise products = c/a, product = -d/a
Symmetric functions	Expressions involving roots that remain unchanged when roots are permuted
Transformation of equations	Changing an equation so that the new roots are a function of the old roots
Reciprocal equation	An equation that remains unchanged when x is replaced by 1/x
Descartes' rule of signs	The number of positive real roots is at most the number of sign changes in the coefficients
Synthetic division	A shortcut method for dividing a polynomial by (x - a) to find remaining roots
Factor theorem	If f(a) = 0, then (x - a) is a factor of f(x)
Remainder theorem	When f(x) is divided by (x - a), the remainder is f(a)
Multiple roots	A root that appears more than once; found using f(x) and f'(x)

Key Formulas

For quadratic equation:

ax² + bx + c = 0

If roots are alpha and beta:

alpha + beta = -b/a
alpha beta = c/a

For cubic equation:

ax³ + bx² + cx + d = 0

If roots are alpha, beta, gamma:

alpha + beta + gamma = -b/a
alpha beta + beta gamma + gamma alpha = c/a
alpha beta gamma = -d/a

I. Choose the Best Answer (MCQs)

1. If the roots of 2x² - 5x + 3 = 0 are alpha and beta, then alpha + beta equals (a) 5/2 (b) -5/2 (c) 3/2 (d) -3/2

Answer(a) 5/2 (By Vieta's formulas, sum of roots = -b/a = -(-5)/2 = 5/2.)

2. The product of the roots of x² - 7x + 12 = 0 is (a) 7 (b) -7 (c) 12 (d) -12

Answer(c) 12 (Product of roots = c/a = 12/1 = 12.)

3. For a cubic equation ax³ + bx² + cx + d = 0, the sum of the pairwise products of roots equals (a) -b/a (b) c/a (c) -d/a (d) d/a

Answer(b) c/a (alpha beta + beta gamma + gamma alpha = c/a by Vieta's formulas for a cubic.)

4. The equation whose roots are 1 and -2 is (a) x² + x - 2 = 0 (b) x² - x - 2 = 0 (c) x² + x + 2 = 0 (d) x² - x + 2 = 0

Answer(b) x² - x - 2 = 0 (Sum = 1 + (-2) = -1; Product = 1 times (-2) = -2; Equation: x² - (sum)x + product = x² + x - 2. Wait — x² - (-1)x + (-2) = x² + x - 2. Actually the standard form is x² - Sx + P = 0 where S = sum, P = product. S = -1, P = -2, so x² - (-1)x + (-2) = x² + x - 2 = 0. Answer is (a).)

Corrected Answer: (a) x² + x - 2 = 0 (Sum of roots S = -1, product P = -2; equation: x² - Sx + P = x² + x - 2.)

5. According to Descartes' rule, the maximum number of positive real roots of x³ - 2x² + x - 1 = 0 is (a) 0 (b) 1 (c) 2 (d) 3

Answer(d) 3 (Sign changes: + to - (1st), - to + (2nd), + to - (3rd) = 3 sign changes, so at most 3 positive real roots.)

6. If alpha is a root of f(x) = 0, then (x - alpha) is a (a) Remainder of f(x) (b) Factor of f(x) (c) Derivative of f(x) (d) Quotient of f(x)

Answer(b) Factor of f(x) (By the Factor Theorem, if f(alpha) = 0, then (x - alpha) is a factor of f(x).)

7. In a reciprocal equation of degree 4, if we substitute y = x + 1/x, the equation reduces to a (a) Linear equation in y (b) Quadratic equation in y (c) Cubic equation in y (d) Quartic equation in y

Answer(b) Quadratic equation in y (Dividing by x² and substituting y = x + 1/x reduces a degree-4 reciprocal equation to a quadratic in y.)

8. If the roots of ax² + bx + c = 0 are in the ratio m : n, then (a) m n b² = a c (m + n)² (b) (m + n)² a c = m n b² (c) m n a c = (m + n)² b² (d) b² m n = a c (m + n)²

Answer(a) m n b² = a c (m + n)² (If roots are m k and n k, then sum = (m+n)k = -b/a and product = mn k² = c/a. Eliminating k gives the relation.)

9. If 2 is a root of x³ - 5x² + 6x = 0, the other roots are (a) 1 and 3 (b) -1 and -3 (c) 0 and 3 (d) 0 and -3

Answer(c) 0 and 3 (Factor out x first: x(x² - 5x + 6) = x(x - 2)(x - 3). So roots are 0, 2, 3.)

10. The number of sign changes in the polynomial f(x) = x⁴ - 3x³ + 2x² + x - 5 is (a) 1 (b) 2 (c) 3 (d) 4

Answer(c) 3 (Signs: +, -, +, +, -. Changes at positions 1→2, 2→3, 4→5. That is 3 sign changes.)

II. Fill in the Blanks

1. For the quadratic equation ax² + bx + c = 0, the sum of roots equals ______.

Answer-b/a

2. The product of the roots of x² - 5x + 6 = 0 is ______.

Answer6

3. For a cubic equation ax³ + bx² + cx + d = 0 with roots alpha, beta, gamma, the product alpha beta gamma equals ______.

Answer-d/a

4. According to the Factor Theorem, if f(k) = 0, then (x - k) is a ______ of f(x).

Answerfactor

5. A reciprocal equation remains unchanged when x is replaced by ______.

Answer1/x

6. If one root of x² - 7x + k = 0 is 3, then k = ______.

Answer12 (Sum = 7, so other root = 4; Product = 3 times 4 = 12, so k = 12.)

7. The remainder when f(x) = x³ - 2x + 1 is divided by (x - 1) is ______.

Answer0 (By the Remainder Theorem, f(1) = 1 - 2 + 1 = 0.)

8. If alpha and beta are roots of x² + px + q = 0, then alpha² + beta² = ______.

Answerp² - 2q (alpha² + beta² = (alpha + beta)² - 2 alpha beta = p² - 2q.)

9. The number of negative real roots of f(x) is found by counting sign changes in ______.

Answerf(-x)

10. An equation of the form ax³ + bx² + bx + a = 0 is called a ______ equation.

Answerreciprocal

III. True or False

1. The sum of the roots of x² + px + q = 0 is p. False. Correction: The sum of the roots is -p (since sum = -b/a = -p/1 = -p).

2. If one root of a quadratic equation is known, the other root can be found using the sum or product of roots. True. (Given one root and the sum (or product), the second root can be computed directly.)

3. A reciprocal equation of odd degree always has x = 1 or x = -1 as a root. True. (For odd-degree reciprocal equations, substituting x = -1 (Type I) or x = 1 (Type II) satisfies the equation.)

4. Descartes' rule of signs gives the exact number of positive real roots. False. Correction: Descartes' rule gives the maximum number of positive real roots. The actual number may be less by an even number.

5. If the discriminant of a quadratic equation is zero, the roots are equal. True. (Discriminant = b² - 4ac = 0 means both roots are real and equal.)

6. Synthetic division can be used only for dividing by linear factors. True. (Synthetic division is a shortcut for dividing a polynomial by (x - a), which is a linear factor.)

7. For a cubic equation, if all three roots are real and positive, the product of roots must be positive. True. (The product alpha beta gamma = -d/a; if all roots are positive, their product is positive, so -d/a > 0.)

8. The substitution y = x + 1/x converts x² + 1/x² into y² - 2. True. (y = x + 1/x implies y² = x² + 2 + 1/x², so x² + 1/x² = y² - 2.)

IV. Match the Following

Column A	Column B
1. Sum of roots (quadratic)	(c) -b/a
2. Product of roots (quadratic)	(e) c/a
3. Product of roots (cubic)	(g) -d/a
4. Factor Theorem	(a) f(k) = 0 implies (x - k) is a factor
5. Remainder Theorem	(f) f(a) is the remainder when f(x) is divided by (x - a)
6. Reciprocal equation	(b) Unchanged when x is replaced by 1/x
7. Descartes' rule	(h) Counts sign changes in coefficients
8. Sum of pairwise products (cubic)	(d) c/a

Answer1-c, 2-e, 3-g, 4-a, 5-f, 6-b, 7-h, 8-d

Standard Answer Pattern

1. Write the given equation. 2. Let the roots be alpha, beta, gamma. 3. Write sum, pairwise product, and product of roots. 4. Use the required transformation. 5. Form the new equation.

Common Question Types

Form an equation whose roots are given.
Find equation with roots alpha + 1, beta + 1.
Find equation with reciprocal roots.
Solve reciprocal equations.
Use synthetic division to find remaining roots.

Worked Pattern

If roots are alpha and beta, equation is:

x² - (sum of roots)x + product of roots = 0

If roots are 2 and 3:

sum = 5, product = 6
equation: x² - 5x + 6 = 0

Worked Examples

Example 1 — Sum and product of roots (Vieta's formulas)

For 2x³ − 3x² + 4x − 5 = 0, find the symmetric functions of the roots.

For ax³ + bx² + cx + d = 0 with roots α, β, γ:

α + β + γ    = −b/a = −(−3)/2 = 3/2
αβ + βγ + γα =  c/a =  4/2     = 2
αβγ          = −d/a = −(−5)/2  = 5/2

Example 2 — Form an equation from its roots

Form the cubic whose roots are 1, 2, and 3.

Σα  = 6,   Σαβ = (1·2)+(2·3)+(3·1) = 11,   αβγ = 6
x³ − (Σα)x² + (Σαβ)x − (αβγ) = 0
→ x³ − 6x² + 11x − 6 = 0

Example 3 — Using a known root to factor

If 1 is a root of x³ − 6x² + 11x − 6 = 0, find the other roots.

x³ − 6x² + 11x − 6 = (x − 1)(x² − 5x + 6) = (x − 1)(x − 2)(x − 3)

The other roots are 2 and 3.

Example 4 — Transformation of equations

If alpha and beta are roots of x² + 5x + 6 = 0, form an equation whose roots are alpha + 1 and beta + 1.

Given: x² + 5x + 6 = 0, so alpha + beta = -5, alpha beta = 6
New roots: alpha + 1 and beta + 1
Sum of new roots = (alpha + 1) + (beta + 1) = alpha + beta + 2 = -5 + 2 = -3
Product of new roots = (alpha + 1)(beta + 1) = alpha beta + alpha + beta + 1 = 6 + (-5) + 1 = 2
New equation: x² - (-3)x + 2 = 0  →  x² + 3x + 2 = 0

Reciprocal Equation Tip

For reciprocal equations, divide by the middle power of x and use:

x + 1/x = y
x² + 1/x² = y² - 2

V. Answer Briefly (2-Mark Questions)

1. State Vieta's formulas for a quadratic equation.

For the quadratic equation ax² + bx + c = 0 with roots alpha and beta, Vieta's formulas state that the sum of roots alpha + beta = -b/a and the product of roots alpha times beta = c/a. These formulas directly connect the coefficients of the equation to the roots without needing to solve the equation explicitly.

2. What is the Factor Theorem?

The Factor Theorem states that if f(k) = 0 for a polynomial f(x), then (x - k) is a factor of f(x). Conversely, if (x - k) is a factor of f(x), then f(k) = 0. This theorem is a special case of the Remainder Theorem and is used to find roots and factorise polynomials efficiently.

3. Write the symmetric functions of the roots of the cubic x³ + px² + qx + r = 0.

If alpha, beta, gamma are the roots of x³ + px² + qx + r = 0, then: alpha + beta + gamma = -p, alpha beta + beta gamma + gamma alpha = q, and alpha beta gamma = -r. These three symmetric functions are the elementary symmetric functions and appear directly in Vieta's formulas for the cubic.

4. What is a reciprocal equation? Give an example.

A reciprocal equation is one that remains unchanged when x is replaced by 1/x. For example, 2x⁴ + 3x³ - 3x - 2 = 0 is a reciprocal equation because substituting 1/x and multiplying by x⁴ gives 2 + 3x - 3x³ - 2x⁴ = 0, which is the original equation with reversed coefficients. Reciprocal equations are solved by substituting y = x + 1/x to reduce the degree.

5. If alpha and beta are roots of x² - 8x + 15 = 0, find alpha² + beta².

We know alpha + beta = 8 and alpha beta = 15. Using the identity alpha² + beta² = (alpha + beta)² - 2(alpha beta) = 64 - 30 = 34. This method avoids finding the individual roots and uses the known symmetric functions directly.

VI. Answer in Detail (5-Mark Questions)

1. Explain the relation between roots and coefficients for a cubic equation and use it to find the symmetric functions of the roots of 2x³ - 4x² + x - 3 = 0.

For a cubic equation ax³ + bx² + cx + d = 0 with roots alpha, beta, gamma, Vieta's formulas give three relations: (i) alpha + beta + gamma = -b/a, (ii) alpha beta + beta gamma + gamma alpha = c/a, (iii) alpha beta gamma = -d/a. These are called the elementary symmetric functions of the roots. For the specific equation 2x³ - 4x² + x - 3 = 0, we have a = 2, b = -4, c = 1, d = -3. Therefore: alpha + beta + gamma = -(-4)/2 = 2, alpha beta + beta gamma + gamma alpha = 1/2, and alpha beta gamma = -(-3)/2 = 3/2. Using these, we can find other symmetric functions like alpha² + beta² + gamma² = (alpha + beta + gamma)² - 2(alpha beta + beta gamma + gamma alpha) = 4 - 1 = 3. Similarly, alpha² beta² + beta² gamma² + gamma² alpha² can be expressed in terms of these elementary symmetric functions.

2. Describe the method of transforming an equation. If alpha and beta are roots of x² + 3x - 10 = 0, form an equation whose roots are 2alpha and 2beta.

Transformation of equations means finding a new equation whose roots are a specified function of the roots of the original equation. The key steps are: write the symmetric functions of the original roots, express the symmetric functions of the new roots in terms of the old ones, and form the new equation. For the given problem, the original equation is x² + 3x - 10 = 0 with alpha + beta = -3 and alpha beta = -10. The new roots are 2alpha and 2beta. The sum of new roots = 2alpha + 2beta = 2(alpha + beta) = 2(-3) = -6. The product of new roots = (2alpha)(2beta) = 4(alpha beta) = 4(-10) = -40. The new equation is x² - (sum)x + (product) = x² + 6x - 40 = 0. An alternative method is to substitute x = y/2 in the original equation: (y/2)² + 3(y/2) - 10 = 0, which gives y²/4 + 3y/2 - 10 = 0, or y² + 6y - 40 = 0.

3. What are reciprocal equations? Explain how to solve a reciprocal equation of degree 4 with an example.

A reciprocal equation is one that remains unchanged when x is replaced by 1/x throughout. There are two types: Type I has symmetric coefficients (coefficients read the same from left to right and right to left), like ax⁴ + bx³ + cx² + bx + a = 0. Type II has anti-symmetric coefficients (coefficients from opposite ends are negatives), like ax⁴ + bx³ + cx² - bx - a = 0. To solve a Type I reciprocal equation of degree 4, divide both sides by x² (since x is not zero): a(x² + 1/x²) + b(x + 1/x) + c = 0. Substitute y = x + 1/x, so x² + 1/x² = y² - 2. This gives a(y² - 2) + by + c = 0, a quadratic in y. Solve for y, then solve x + 1/x = y for each value of y to get x² - yx + 1 = 0. For example, solving x⁴ - 5x³ + 6x² - 5x + 1 = 0: divide by x² to get x² - 5x + 6 - 5/x + 1/x² = 0, which rearranges to (x² + 1/x²) - 5(x + 1/x) + 6 = 0. Let y = x + 1/x, so y² - 2 - 5y + 6 = 0, i.e., y² - 5y + 4 = 0. This gives y = 1 or y = 4. Then x + 1/x = 1 gives x² - x + 1 = 0 (complex roots), and x + 1/x = 4 gives x² - 4x + 1 = 0 (x = 2 plus or minus sqrt(3)).

4. State and explain Descartes' rule of signs. Apply it to determine the possible number of positive and negative real roots of f(x) = x⁴ - 3x³ + 2x² + x - 1.

Descartes' rule of signs states that the number of positive real roots of a polynomial f(x) is at most equal to the number of sign changes in the sequence of coefficients of f(x), and differs from that number by an even integer. Similarly, the number of negative real roots is at most equal to the number of sign changes in the coefficients of f(-x). For f(x) = x⁴ - 3x³ + 2x² + x - 1, the coefficients are +1, -3, +2, +1, -1. Sign changes: +1 to -3 (1st change), -3 to +2 (2nd change), +2 to +1 (no change), +1 to -1 (3rd change). There are 3 sign changes, so the number of positive real roots is at most 3 (possible: 3 or 1). For negative roots, compute f(-x) = x⁴ + 3x³ + 2x² - x - 1. Coefficients: +1, +3, +2, -1, -1. Sign changes: +2 to -1 (1st change). There is 1 sign change, so there is exactly 1 negative real root.

Common Mistakes to Avoid

1. ❌ Writing sum of roots as b/a instead of -b/a. ✅ Always use the formula: sum of roots = -b/a. The negative sign is essential and a very common source of error.

2. ❌ Forgetting the negative sign in the product of roots of a cubic (alpha beta gamma = -d/a). ✅ For a cubic ax³ + bx² + cx + d = 0, the product of roots is -d/a, not d/a. The signs alternate.

3. ❌ Confusing transformation substitution direction. ✅ To form an equation with roots (alpha + k), substitute x = y - k in the original equation. The substitution goes opposite to the transformation.

4. ❌ Not checking whether x = 0 is a root before dividing by xⁿ in reciprocal equations. ✅ Always verify that x = 0 is not a root before dividing by a power of x, since division by x assumes x is not zero.

5. ❌ Applying Descartes' rule directly for negative roots without computing f(-x). ✅ For negative roots, always compute f(-x) first, then count sign changes in the resulting polynomial.

6. ❌ Assuming Descartes' rule gives the exact number of roots. ✅ Descartes' rule gives only the maximum number of positive/negative real roots. The actual number may be less by a multiple of 2.

7. ❌ Making sign errors while expanding (alpha + k)(beta + k) in transformation problems. ✅ Use (alpha + k)(beta + k) = alpha beta + k(alpha + beta) + k². Compute step by step using known values.

8. ❌ Confusing reciprocal equation types (symmetric vs. anti-symmetric coefficients). ✅ Type I: coefficients read the same forwards and backwards (e.g., ax⁴ + bx³ + cx² + bx + a). Type II: coefficients from opposite ends are negatives (e.g., ax⁴ + bx³ + cx² - bx - a).

Exam Checklist

Write the Vieta's formulas for the given degree of equation at the start of every problem.
Use consistent symbols for roots (alpha, beta, gamma) throughout the solution.
Check signs carefully: sum = -b/a, product of pairwise = c/a, product = -d/a for a cubic.
For transformation problems, substitute correctly (opposite direction to the transformation).
Verify reciprocal equation type before applying the substitution y = x + 1/x.
Factor out x if x = 0 could be a root before dividing by xⁿ.
Count sign changes carefully when applying Descartes' rule.
State whether the number of roots is exact or an upper bound when using Descartes' rule.
Use synthetic division for finding remaining roots efficiently.
Express the final answer as a simplified polynomial equation equal to zero.
Double-check that the degree of the new equation matches the expected degree after transformation.

Frequently Asked Questions

Memorise the relation between roots and coefficients for quadratic and cubic equations, then practise transformation and reciprocal equation problems.