Samacheer Kalvi Class 12 Maths Solutions Chapter 3 Theory of Equations Exercise 3.3

Question 1. Solve the cubic equation: \( 2x^3 - x^2 - 18x + 9 = 0 \) if sum of two of its roots vanishes.
Answer: We are given the cubic equation: \( 2x^3 - x^2 - 18x + 9 = 0 \). To make calculations simpler, divide the entire equation by 2:
\( x^3 - \frac{x^2}{2} - 9x + \frac{9}{2} = 0 \) Let the three roots of this equation be \( \alpha, -\alpha, \beta \). The problem states that the sum of two roots is zero, which means one root is the negative of another. From Vieta's formulas: Sum of the roots: \( \alpha + (-\alpha) + \beta = -\left(\frac{-1}{2}\right) \)
\( \implies \beta = \frac{1}{2} \) Product of the roots: \( (\alpha)(-\alpha)(\beta) = -\frac{9}{2} \)
\( \implies -\alpha^2 \left(\frac{1}{2}\right) = -\frac{9}{2} \)
\( \implies -\alpha^2 = -9 \)
\( \implies \alpha^2 = 9 \)
\( \implies \alpha = \pm 3 \) So, the possible values for \( \alpha \) are 3 and -3. The three roots are \( \alpha, -\alpha, \beta \). If \( \alpha = 3 \), then \( -\alpha = -3 \). Since \( \beta = \frac{1}{2} \), the roots are \( 3, -3, \frac{1}{2} \). These roots satisfy the condition that two of them sum to zero (\( 3 + (-3) = 0 \)).In simple words: We find the values of \( \alpha \) and \( \beta \) using special rules about how roots of equations are related. We discovered that two roots are opposites of each other (like 3 and -3), and the third root is \( \frac{1}{2} \). These three numbers are the solutions to the equation.

🎯 Exam Tip: When given conditions about roots (like sum or product), always use Vieta's formulas to set up equations. This is the most efficient way to find the roots or coefficients.

Question 2. Solve the equation \( 9x^3 - 36x^2 + 44x - 16 = 0 \) if the roots form an arithmetic progression.
Answer: We have the equation: \( 9x^3 - 36x^2 + 44x - 16 = 0 \). Since the roots form an arithmetic progression (AP), we can represent them as \( a - d, a, a + d \). First, divide the entire equation by 9 to simplify it, making the coefficient of \( x^3 \) equal to 1:
\( x^3 - 4x^2 + \frac{44}{9}x - \frac{16}{9} = 0 \) Now, using Vieta's formulas: Sum of the roots: \( (a - d) + a + (a + d) = -(-4) \)
\( \implies 3a = 4 \)
\( \implies a = \frac{4}{3} \) Product of the roots: \( (a - d)(a)(a + d) = -(-\frac{16}{9}) \)
\( \implies a(a^2 - d^2) = \frac{16}{9} \) Substitute the value of \( a = \frac{4}{3} \) into this equation:
\( \frac{4}{3} \left( \left(\frac{4}{3}\right)^2 - d^2 \right) = \frac{16}{9} \)
\( \implies \frac{4}{3} \left( \frac{16}{9} - d^2 \right) = \frac{16}{9} \) Divide both sides by \( \frac{4}{3} \):
\( \frac{16}{9} - d^2 = \frac{16}{9} \div \frac{4}{3} \)
\( \implies \frac{16}{9} - d^2 = \frac{16}{9} \times \frac{3}{4} \)
\( \implies \frac{16}{9} - d^2 = \frac{4}{3} \)
\( \implies d^2 = \frac{16}{9} - \frac{4}{3} \)
\( \implies d^2 = \frac{16}{9} - \frac{12}{9} \)
\( \implies d^2 = \frac{4}{9} \)
\( \implies d = \pm \sqrt{\frac{4}{9}} \)
\( \implies d = \pm \frac{2}{3} \) We can choose \( d = \frac{2}{3} \) (choosing \( d = -\frac{2}{3} \) would just reverse the order of the roots). Now find the roots: First root: \( a - d = \frac{4}{3} - \frac{2}{3} = \frac{2}{3} \) Second root: \( a = \frac{4}{3} \) Third root: \( a + d = \frac{4}{3} + \frac{2}{3} = \frac{6}{3} = 2 \) So, the roots of the equation are \( \frac{2}{3}, \frac{4}{3}, 2 \). These three numbers increase by the same amount each time, showing they are in an arithmetic progression.In simple words: We used the idea that the roots are in an arithmetic progression, which means they go up by a steady amount. By using special rules about adding and multiplying roots, we found the middle root first. Then, we found the common difference between the roots. This helped us figure out all three roots of the equation.

🎯 Exam Tip: When roots are in AP, always assume them as \( a-d, a, a+d \). For GP, assume them as \( \frac{a}{\lambda}, a, a\lambda \). This simplifies the sum/product calculations considerably.

Question 3. Solve the equation \( 3x^3 - 26x^2 + 52x - 24 = 0 \) if its roots form a geometric progression.
Answer: We have the equation: \( 3x^3 - 26x^2 + 52x - 24 = 0 \). Since the roots form a geometric progression (GP), we can represent them as \( \frac{a}{\lambda}, a, a\lambda \). First, divide the entire equation by 3 to make the coefficient of \( x^3 \) equal to 1:
\( x^3 - \frac{26}{3}x^2 + \frac{52}{3}x - \frac{24}{3} = 0 \)
\( \implies x^3 - \frac{26}{3}x^2 + \frac{52}{3}x - 8 = 0 \) Now, using Vieta's formulas: Product of the roots: \( \left(\frac{a}{\lambda}\right)(a)(a\lambda) = -(-8) \)
\( \implies a^3 = 8 \)
\( \implies a = \sqrt[3]{8} \)
\( \implies a = 2 \) Sum of the roots: \( \frac{a}{\lambda} + a + a\lambda = -(-\frac{26}{3}) \)
\( \implies a \left(\frac{1}{\lambda} + 1 + \lambda\right) = \frac{26}{3} \) Substitute \( a = 2 \):
\( 2 \left(\frac{1 + \lambda + \lambda^2}{\lambda}\right) = \frac{26}{3} \)
\( \implies \frac{1 + \lambda + \lambda^2}{\lambda} = \frac{26}{3 \times 2} \)
\( \implies \frac{1 + \lambda + \lambda^2}{\lambda} = \frac{13}{3} \)
\( \implies 3(1 + \lambda + \lambda^2) = 13\lambda \)
\( \implies 3 + 3\lambda + 3\lambda^2 = 13\lambda \)
\( \implies 3\lambda^2 + 3\lambda - 13\lambda + 3 = 0 \)
\( \implies 3\lambda^2 - 10\lambda + 3 = 0 \) This is a quadratic equation in \( \lambda \). We can factor it: We need two numbers that multiply to \( 3 \times 3 = 9 \) and add to -10. These numbers are -9 and -1. \( 3\lambda^2 - 9\lambda - \lambda + 3 = 0 \)
\( \implies 3\lambda(\lambda - 3) - 1(\lambda - 3) = 0 \)
\( \implies (3\lambda - 1)(\lambda - 3) = 0 \) So, \( 3\lambda - 1 = 0 \) or \( \lambda - 3 = 0 \). This means \( \lambda = \frac{1}{3} \) or \( \lambda = 3 \). Now, we find the roots using \( a=2 \) and both values of \( \lambda \): Case 1: If \( \lambda = 3 \) The roots are \( \frac{a}{\lambda}, a, a\lambda \). \( \frac{2}{3}, 2, 2 \times 3 = 6 \) The roots are \( \frac{2}{3}, 2, 6 \). Case 2: If \( \lambda = \frac{1}{3} \) The roots are \( \frac{a}{\lambda}, a, a\lambda \). \( \frac{2}{1/3}, 2, 2 \times \frac{1}{3} \) \( 2 \times 3, 2, \frac{2}{3} \) \( 6, 2, \frac{2}{3} \) Both cases give the same set of roots, just in a different order. The roots of the equation are \( \frac{2}{3}, 2, 6 \). We notice that each number is three times the previous one (or one-third of the next), showing they are in a geometric progression.In simple words: Since the problem said the roots form a geometric progression, we used special formulas based on multiplying and adding roots. We first found the middle root. Then, we solved for the common ratio between the roots, which allowed us to find all three roots.

🎯 Exam Tip: For geometric progressions, always calculate the product of roots first as it often directly gives you 'a' (the middle term), simplifying the process significantly.

Question 4. Determine k and solve the equation \( 2x^3 - 6x^2 + 3x + k = 0 \) if one of its roots is twice the sum of the other two roots.
Answer: We are given the cubic equation: \( 2x^3 - 6x^2 + 3x + k = 0 \). Let the three roots be \( \alpha, \beta, \gamma \). The problem states that one root is twice the sum of the other two. Let's assume \( \alpha = 2(\beta + \gamma) \). (Equation 1) First, divide the entire equation by 2 to make the coefficient of \( x^3 \) equal to 1:
\( x^3 - 3x^2 + \frac{3}{2}x + \frac{k}{2} = 0 \) Now, using Vieta's formulas: Sum of the roots: \( \alpha + \beta + \gamma = -(-3) \)
\( \implies \alpha + \beta + \gamma = 3 \) (Equation A) Substitute Equation 1 into Equation A:
\( 2(\beta + \gamma) + (\beta + \gamma) = 3 \)
\( \implies 3(\beta + \gamma) = 3 \)
\( \implies \beta + \gamma = 1 \) Now substitute \( \beta + \gamma = 1 \) back into Equation 1:
\( \alpha = 2(1) \)
\( \implies \alpha = 2 \) So, one root is \( \alpha = 2 \). Since \( \alpha \) is a root, substituting \( x=2 \) into the original equation should make the equation true. This helps us find k. \( 2(2)^3 - 6(2)^2 + 3(2) + k = 0 \)
\( \implies 2(8) - 6(4) + 6 + k = 0 \)
\( \implies 16 - 24 + 6 + k = 0 \)
\( \implies -2 + k = 0 \)
\( \implies k = 2 \) So, the value of \( k \) is 2. Now we need to solve the equation with \( k=2 \):
\( 2x^3 - 6x^2 + 3x + 2 = 0 \) Since we know \( x = 2 \) is a root, \( (x - 2) \) must be a factor of the polynomial. We can use synthetic division to find the other factors. \[ \begin{array}{c|cccc} 2 & 2 & -6 & 3 & 2 \\ & & 4 & -4 & -2 \\ \hline & 2 & -2 & -1 & 0 \\ \end{array} \] The quotient is \( 2x^2 - 2x - 1 = 0 \). Now we solve this quadratic equation using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) Here, \( a=2, b=-2, c=-1 \). \( x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(2)(-1)}}{2(2)} \)
\( \implies x = \frac{2 \pm \sqrt{4 + 8}}{4} \)
\( \implies x = \frac{2 \pm \sqrt{12}}{4} \)
\( \implies x = \frac{2 \pm 2\sqrt{3}}{4} \)
\( \implies x = \frac{2(1 \pm \sqrt{3})}{4} \)
\( \implies x = \frac{1 \pm \sqrt{3}}{2} \) So the other two roots are \( \frac{1 + \sqrt{3}}{2} \) and \( \frac{1 - \sqrt{3}}{2} \). The roots of the equation are \( 2, \frac{1 + \sqrt{3}}{2}, \frac{1 - \sqrt{3}}{2} \). We confirmed the value of k by using the known root, which is a great way to verify the solution.In simple words: We used the given condition about one root being double the sum of the others to find one root, which was 2. Then, we used this root to find the value of 'k' in the equation. Once 'k' was known, we divided the main equation by \( (x-2) \) and solved the remaining quadratic equation to find the other two roots.

🎯 Exam Tip: If one root of a polynomial is known, always use synthetic division to reduce the degree of the polynomial, making it easier to find the remaining roots.

Question 5. Find all zeros of the polynomial \( x^6 - 3x^5 - 5x^4 + 22x^3 - 39x^2 - 39x + 135 = 0 \), if it is known that \( 1 + 2i \) and \( \sqrt{3} \) are two of its zeros.
Answer: We are given the polynomial \( P(x) = x^6 - 3x^5 - 5x^4 + 22x^3 - 39x^2 - 39x + 135 \). We know that if a polynomial has real coefficients (which this one does), then complex roots always come in conjugate pairs, and irrational roots also come in conjugate pairs (if the equation is formed from rational coefficients, which is usually the case unless specified). Given zeros: 1. \( 1 + 2i \) 2. \( \sqrt{3} \) From these, we can identify two more zeros: 3. The conjugate of \( 1 + 2i \) is \( 1 - 2i \). 4. The conjugate of \( \sqrt{3} \) is \( -\sqrt{3} \). Now we have four zeros: \( 1+2i, 1-2i, \sqrt{3}, -\sqrt{3} \). Let's find the quadratic factor for each pair of conjugate roots: For \( 1 + 2i \) and \( 1 - 2i \): Sum of roots \( = (1 + 2i) + (1 - 2i) = 2 \) Product of roots \( = (1 + 2i)(1 - 2i) = 1^2 - (2i)^2 = 1 - 4i^2 = 1 - 4(-1) = 1 + 4 = 5 \) The quadratic factor is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = x^2 - 2x + 5 \). For \( \sqrt{3} \) and \( -\sqrt{3} \): Sum of roots \( = \sqrt{3} + (-\sqrt{3}) = 0 \) Product of roots \( = (\sqrt{3})(-\sqrt{3}) = -3 \) The quadratic factor is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = x^2 - 0x - 3 = x^2 - 3 \). Now we have two quadratic factors: \( (x^2 - 2x + 5) \) and \( (x^2 - 3) \). Multiplying these factors gives a polynomial of degree 4: \( (x^2 - 2x + 5)(x^2 - 3) = x^2(x^2 - 3) - 2x(x^2 - 3) + 5(x^2 - 3) \) \( = x^4 - 3x^2 - 2x^3 + 6x + 5x^2 - 15 \) \( = x^4 - 2x^3 + 2x^2 + 6x - 15 \) This is a factor of the given polynomial \( P(x) \). We can perform polynomial long division or compare coefficients. Given the complexity of the polynomial, comparing coefficients is often easier. Let the original polynomial be \( P(x) = (x^4 - 2x^3 + 2x^2 + 6x - 15)(x^2 + px + q) \). Comparing the leading terms: \( x^6 = x^4 \cdot x^2 \), which is correct. Comparing the constant terms: \( 135 = (-15) \cdot q \)
\( \implies q = \frac{135}{-15} \)
\( \implies q = -9 \) So the other factor is of the form \( (x^2 + px - 9) \). Now we compare the coefficients of \( x \): The coefficient of \( x \) in \( P(x) \) is \( -39 \). From \( (x^4 - 2x^3 + 2x^2 + 6x - 15)(x^2 + px - 9) \), the terms contributing to \( x \) are: \( (6x)(q) + (-15)(px) \) \( 6x(-9) + (-15)(px) = -54x - 15px \) So, \( -54 - 15p = -39 \)
\( \implies -15p = -39 + 54 \)
\( \implies -15p = 15 \)
\( \implies p = -1 \) The remaining quadratic factor is \( x^2 - x - 9 \). To find the remaining zeros, we solve \( x^2 - x - 9 = 0 \) using the quadratic formula: \( x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-9)}}{2(1)} \)
\( \implies x = \frac{1 \pm \sqrt{1 + 36}}{2} \)
\( \implies x = \frac{1 \pm \sqrt{37}}{2} \) So, the other two zeros are \( \frac{1 + \sqrt{37}}{2} \) and \( \frac{1 - \sqrt{37}}{2} \). The six zeros of the polynomial are \( 1 + 2i, 1 - 2i, \sqrt{3}, -\sqrt{3}, \frac{1 + \sqrt{37}}{2}, \frac{1 - \sqrt{37}}{2} \). Finding factors from known roots is a key strategy for solving higher-degree polynomials.In simple words: We started with two given roots and used the rule that complex and irrational roots come in pairs. This gave us four roots. We then built a smaller polynomial from these four roots. By comparing our new polynomial with the original one, we found the last part of the polynomial. Solving this last part gave us the final two roots.

🎯 Exam Tip: Always remember that for polynomials with real coefficients, complex roots (\( a+bi \)) always appear with their conjugates (\( a-bi \)), and irrational roots (\( a+\sqrt{b} \)) with their conjugates (\( a-\sqrt{b} \)). Use this property to find additional roots immediately.

Question 6. Solve the cubic equations:
(i) \( 2x^3 - 9x^2 + 10x - 3 = 0 \)
(ii) \( 8x^3 - 2x^2 - 7x + 3 = 0 \)
Answer:
(i) For the equation \( 2x^3 - 9x^2 + 10x - 3 = 0 \): First, check for simple integer roots by summing coefficients. Sum of coefficients: \( 2 - 9 + 10 - 3 = 0 \). Since the sum of the coefficients is 0, \( x = 1 \) is a root. This means \( (x - 1) \) is a factor. We use synthetic division to find the other factor. \[ \begin{array}{c|cccc} 1 & 2 & -9 & 10 & -3 \\ & & 2 & -7 & 3 \\ \hline & 2 & -7 & 3 & 0 \\ \end{array} \] The quotient is \( 2x^2 - 7x + 3 = 0 \). Now solve this quadratic equation. We can factor it: We need two numbers that multiply to \( 2 \times 3 = 6 \) and add to -7. These are -6 and -1. \( 2x^2 - 6x - x + 3 = 0 \)
\( \implies 2x(x - 3) - 1(x - 3) = 0 \)
\( \implies (2x - 1)(x - 3) = 0 \) So, \( 2x - 1 = 0 \) or \( x - 3 = 0 \). This means \( x = \frac{1}{2} \) or \( x = 3 \). The roots of the equation are \( 1, \frac{1}{2}, 3 \).
(ii) For the equation \( 8x^3 - 2x^2 - 7x + 3 = 0 \): Let's check if \( x = 1 \) or \( x = -1 \) are roots. Sum of coefficients: \( 8 - 2 - 7 + 3 = 2 \neq 0 \), so \( x = 1 \) is not a root. Sum of coefficients of odd powers (8x³ - 7x): \( 8 - 7 = 1 \) Sum of coefficients of even powers ( -2x² + 3): \( -2 + 3 = 1 \) Since the sums are equal, \( x = -1 \) is a root. This means \( (x + 1) \) is a factor. We use synthetic division. \[ \begin{array}{c|cccc} -1 & 8 & -2 & -7 & 3 \\ & & -8 & 10 & -3 \\ \hline & 8 & -10 & 3 & 0 \\ \end{array} \] The quotient is \( 8x^2 - 10x + 3 = 0 \). Now solve this quadratic equation. We can factor it: We need two numbers that multiply to \( 8 \times 3 = 24 \) and add to -10. These are -6 and -4. \( 8x^2 - 4x - 6x + 3 = 0 \)
\( \implies 4x(2x - 1) - 3(2x - 1) = 0 \)
\( \implies (4x - 3)(2x - 1) = 0 \) So, \( 4x - 3 = 0 \) or \( 2x - 1 = 0 \). This means \( x = \frac{3}{4} \) or \( x = \frac{1}{2} \). The roots of the equation are \( -1, \frac{3}{4}, \frac{1}{2} \). Checking for simple integer roots first can save a lot of time.In simple words: For both equations, we first tried to find simple roots like 1 or -1 by checking the sum of coefficients. If we found a root, we used synthetic division to break down the cubic equation into a simpler quadratic equation. Then, we solved the quadratic part to find the remaining roots.

🎯 Exam Tip: Always check if \( x=1 \) (sum of all coefficients is 0) or \( x=-1 \) (sum of coefficients of even powers equals sum of coefficients of odd powers) are roots for cubic equations. This can quickly provide one root, simplifying the problem to a quadratic equation.

Question 7. Solve the equation: \( x^4 - 14x^2 + 45 = 0 \)
Answer: We have the equation: \( x^4 - 14x^2 + 45 = 0 \). This is a biquadratic equation. We can solve it by making a substitution. Let \( t = x^2 \). Then \( x^4 = (x^2)^2 = t^2 \). Substituting \( t \) into the equation:
\( t^2 - 14t + 45 = 0 \) This is a quadratic equation in \( t \). We can factor it. We need two numbers that multiply to 45 and add to -14. These numbers are -9 and -5. \( t^2 - 9t - 5t + 45 = 0 \)
\( \implies t(t - 9) - 5(t - 9) = 0 \)
\( \implies (t - 9)(t - 5) = 0 \) So, \( t - 9 = 0 \) or \( t - 5 = 0 \). This means \( t = 9 \) or \( t = 5 \). Now substitute back \( x^2 \) for \( t \): Case 1: \( x^2 = 9 \) Taking the square root of both sides: \( x = \pm \sqrt{9} \)
\( \implies x = \pm 3 \) Case 2: \( x^2 = 5 \) Taking the square root of both sides: \( x = \pm \sqrt{5} \) Therefore, the four roots of the equation are \( 3, -3, \sqrt{5}, -\sqrt{5} \). Substituting a variable like 't' often simplifies equations that look complex at first glance.In simple words: We changed the complex-looking equation into a simpler quadratic equation by replacing \( x^2 \) with a new variable, 't'. We solved for 't', and then put \( x^2 \) back in to find the actual values of \( x \). This gave us the four solutions for the original equation.

🎯 Exam Tip: For equations of the form \( ax^{2n} + bx^n + c = 0 \), always use the substitution \( t = x^n \) to reduce it to a quadratic equation \( at^2 + bt + c = 0 \), which is much easier to solve.