Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.14

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.14

Question 1. Write each of the following expression in terms of \( \alpha + \beta \) and \( \alpha\beta \)
(i) \( \frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha} \)
(ii) \( \frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha} \)
(iii) \( (3\alpha – 1) (3\beta – 1) \)
(iv) \( \frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha} \)
Answer:
(i) \( \frac{\alpha}{3 \beta}+\frac{\beta}{3 \alpha} = \frac{\alpha^2+\beta^2}{3\alpha\beta} = \frac{(\alpha+\beta)^2-2\alpha\beta}{3\alpha\beta} \) This uses a common algebraic identity to simplify the expression.
(ii) \( \frac{1}{\alpha^{2} \beta}+\frac{1}{\beta^{2} \alpha} = \frac{\beta+\alpha}{\alpha^{2} \beta^{2}} = \frac{\alpha+\beta}{(\alpha\beta)^{2}} \) We find a common denominator to combine the fractions.
(iii) \( (3\alpha – 1) (3\beta – 1) = 9\alpha\beta – 3\alpha – 3\beta + 1 = 9\alpha\beta – 3(\alpha + \beta) + 1 \) We expand the product and group common terms.
(iv) \( \frac{\alpha+3}{\beta}+\frac{\beta+3}{\alpha} = \frac{\alpha(\alpha+3)+\beta(\beta+3)}{\alpha\beta} \)
\( \implies \frac{\alpha^2+3\alpha+\beta^2+3\beta}{\alpha\beta} \)
\( \implies \frac{\alpha^2+\beta^2+3(\alpha+\beta)}{\alpha\beta} \)
\( \implies \frac{(\alpha+\beta)^2 - 2\alpha\beta + 3(\alpha+\beta)}{\alpha\beta} \) This step converts the sum of squares into a form involving `\(\alpha+\beta\)` and `\(\alpha\beta\)`.
Using values (from question 2 context if applicable) \( \alpha+\beta = \frac{7}{2} \) and \( \alpha\beta = \frac{5}{2} \):
\( \implies \frac{(\frac{7}{2})^2 - 2(\frac{5}{2}) + 3(\frac{7}{2})}{\frac{5}{2}} \)
\( \implies \frac{\frac{49}{4} - 5 + \frac{21}{2}}{\frac{5}{2}} \)
\( \implies \frac{\frac{49}{4} - \frac{20}{4} + \frac{42}{4}}{\frac{5}{2}} \)
\( \implies \frac{\frac{49 - 20 + 42}{4}}{\frac{5}{2}} \)
\( \implies \frac{\frac{71}{4}}{\frac{5}{2}} = \frac{71}{4} \times \frac{2}{5} = \frac{71}{10} \) Here, we simplify the complex fraction by multiplying by the reciprocal.
In simple words: For each expression, we change it so it only uses the sum of the roots (`\(\alpha + \beta\)`) and the product of the roots (`\(\alpha\beta\)`). This is done using basic algebra and identities to combine terms or find common denominators.

🎯 Exam Tip: Remember key algebraic identities like \( (\alpha+\beta)^2 = \alpha^2+\beta^2+2\alpha\beta \) and \( (\alpha-\beta)^2 = (\alpha+\beta)^2-4\alpha\beta \) as they are frequently used in these types of problems.

Question 2. The roots of the equation \( 2x^2 – 7x + 5 = 0 \) are \( \alpha \) and \( \beta \). Find the value of [without solving the equation]
(i) \( \frac{1}{\alpha}+\frac{1}{\beta} \)
(ii) \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} \)
(iii) \( \frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2} \)
Answer:
For the equation \( 2x^2 – 7x + 5 = 0 \), we have \( a=2, b=-7, c=5 \).
Sum of the roots: \( \alpha + \beta = -\frac{b}{a} = -\frac{-7}{2} = \frac{7}{2} \)
Product of the roots: \( \alpha\beta = \frac{c}{a} = \frac{5}{2} \)

(i) \( \frac{1}{\alpha}+\frac{1}{\beta} = \frac{\beta+\alpha}{\alpha\beta} \)
\( \implies \frac{\frac{7}{2}}{\frac{5}{2}} = \frac{7}{2} \times \frac{2}{5} = \frac{7}{5} \) We substitute the values of the sum and product of roots.

(ii) \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} \)
We know \( \alpha^2+\beta^2 = (\alpha+\beta)^2-2\alpha\beta \)
So, \( \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{(\alpha+\beta)^2-2\alpha\beta}{\alpha\beta} \)
\( \implies \frac{(\frac{7}{2})^2 - 2(\frac{5}{2})}{\frac{5}{2}} \)
\( \implies \frac{\frac{49}{4} - 5}{\frac{5}{2}} \)
\( \implies \frac{\frac{49-20}{4}}{\frac{5}{2}} \)
\( \implies \frac{\frac{29}{4}}{\frac{5}{2}} = \frac{29}{4} \times \frac{2}{5} = \frac{29}{10} \) This evaluates the expression by plugging in the calculated values.

(iii) \( \frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2} = \frac{(\alpha+2)^2+(\beta+2)^2}{(\alpha+2)(\beta+2)} \)
Numerator: \( (\alpha+2)^2+(\beta+2)^2 = (\alpha^2+4\alpha+4) + (\beta^2+4\beta+4) = \alpha^2+\beta^2+4(\alpha+\beta)+8 \)
\( \implies (\alpha+\beta)^2-2\alpha\beta+4(\alpha+\beta)+8 \)
Denominator: \( (\alpha+2)(\beta+2) = \alpha\beta+2\alpha+2\beta+4 = \alpha\beta+2(\alpha+\beta)+4 \)
Now, substitute the values \( \alpha+\beta = \frac{7}{2} \) and \( \alpha\beta = \frac{5}{2} \):
Numerator: \( (\frac{7}{2})^2 - 2(\frac{5}{2}) + 4(\frac{7}{2}) + 8 = \frac{49}{4} - 5 + 14 + 8 = \frac{49}{4} - 5 + 22 = \frac{49}{4} + 17 = \frac{49+68}{4} = \frac{117}{4} \)
Denominator: \( \frac{5}{2} + 2(\frac{7}{2}) + 4 = \frac{5}{2} + 7 + 4 = \frac{5}{2} + 11 = \frac{5+22}{2} = \frac{27}{2} \)
So, \( \frac{\alpha+2}{\beta+2}+\frac{\beta+2}{\alpha+2} = \frac{\frac{117}{4}}{\frac{27}{2}} = \frac{117}{4} \times \frac{2}{27} = \frac{117}{2 \times 27} = \frac{117}{54} \) We perform the final division to get the simplified answer.
In simple words: First, we find the sum and product of the roots of the given equation. Then, for each part, we change the expression to use these sum and product values. Finally, we put in the numbers and solve.

🎯 Exam Tip: Always correctly identify 'a', 'b', and 'c' from the quadratic equation \( ax^2+bx+c=0 \) to correctly calculate \( \alpha+\beta \) and \( \alpha\beta \).

Question 3. The roots of the equation \( x^2 + 6x – 4 = 0 \) are \( \alpha, \beta \). Find the quadratic equation whose roots are
(i) \( \alpha^2 \) and \( \beta^2 \)
(ii) \( \frac{2}{\alpha} \) and \( \frac{2}{\beta} \)
(iii) \( \alpha^2\beta \) and \( \beta^2\alpha \)
Answer:
For the equation \( x^2 + 6x – 4 = 0 \), we have \( a=1, b=6, c=-4 \).
Sum of the roots: \( \alpha + \beta = -\frac{b}{a} = -\frac{6}{1} = -6 \)
Product of the roots: \( \alpha\beta = \frac{c}{a} = \frac{-4}{1} = -4 \)

(i) Roots are \( \alpha^2 \) and \( \beta^2 \).
Sum of new roots: \( \alpha^2 + \beta^2 = (\alpha+\beta)^2-2\alpha\beta = (-6)^2 - 2(-4) = 36 + 8 = 44 \)
Product of new roots: \( \alpha^2\beta^2 = (\alpha\beta)^2 = (-4)^2 = 16 \)
The quadratic equation is \( x^2 - (\text{Sum of the roots})x + (\text{Product of the roots}) = 0 \)
\( \implies x^2 - (44)x + 16 = 0 \)
\( \implies x^2 - 44x + 16 = 0 \) This is the required quadratic equation.

(ii) Roots are \( \frac{2}{\alpha} \) and \( \frac{2}{\beta} \).
Sum of new roots: \( \frac{2}{\alpha}+\frac{2}{\beta} = \frac{2\beta+2\alpha}{\alpha\beta} = \frac{2(\alpha+\beta)}{\alpha\beta} \)
\( \implies \frac{2(-6)}{-4} = \frac{-12}{-4} = 3 \)
Product of new roots: \( \frac{2}{\alpha} \times \frac{2}{\beta} = \frac{4}{\alpha\beta} \)
\( \implies \frac{4}{-4} = -1 \)
The quadratic equation is \( x^2 - (\text{Sum of the roots})x + (\text{Product of the roots}) = 0 \)
\( \implies x^2 - (3)x + (-1) = 0 \)
\( \implies x^2 - 3x - 1 = 0 \) This is the required quadratic equation.

(iii) Roots are \( \alpha^2\beta \) and \( \beta^2\alpha \).
Sum of new roots: \( \alpha^2\beta + \beta^2\alpha = \alpha\beta(\alpha+\beta) \)
\( \implies (-4)(-6) = 24 \)
Product of new roots: \( \alpha^2\beta \times \beta^2\alpha = \alpha^3\beta^3 = (\alpha\beta)^3 \)
\( \implies (-4)^3 = -64 \)
The quadratic equation is \( x^2 - (\text{Sum of the roots})x + (\text{Product of the roots}) = 0 \)
\( \implies x^2 - (24)x + (-64) = 0 \)
\( \implies x^2 - 24x - 64 = 0 \) This is the required quadratic equation.
In simple words: For each part, we first find the sum and product of the new roots using the `\(\alpha+\beta\)` and `\(\alpha\beta\)` values from the original equation. Then, we use the general formula `\(x^2 - (\text{sum})x + (\text{product}) = 0\)` to create the new quadratic equation.

🎯 Exam Tip: Remember that if \( r_1 \) and \( r_2 \) are the roots of a quadratic equation, the equation itself is \( x^2 - (r_1+r_2)x + r_1r_2 = 0 \).

Question 4. If \( \alpha, \beta \) are the roots of \( 7x^2 + ax + 2 = 0 \) and if \( \beta - \alpha = \frac{13}{7} \) Find the values of a.
Answer:
For the equation \( 7x^2 + ax + 2 = 0 \), we have \( A=7, B=a, C=2 \).
Sum of the roots: \( \alpha + \beta = -\frac{B}{A} = -\frac{a}{7} \)
Product of the roots: \( \alpha\beta = \frac{C}{A} = \frac{2}{7} \)
Given: \( \beta - \alpha = \frac{13}{7} \). This means \( \alpha - \beta = -\frac{13}{7} \).
We know the identity: \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \)
Substitute the known values:
\( (-\frac{13}{7})^2 = (-\frac{a}{7})^2 - 4(\frac{2}{7}) \)
\( \implies \frac{169}{49} = \frac{a^2}{49} - \frac{8}{7} \)
\( \implies \frac{169}{49} = \frac{a^2}{49} - \frac{8 \times 7}{7 \times 7} \)
\( \implies \frac{169}{49} = \frac{a^2}{49} - \frac{56}{49} \)
Multiply by 49 to clear denominators:
\( \implies 169 = a^2 - 56 \)
\( \implies a^2 = 169 + 56 \)
\( \implies a^2 = 225 \)
\( \implies a = \pm\sqrt{225} \)
\( \implies a = \pm 15 \) So, 'a' can be either 15 or -15.
In simple words: We first write down the sum and product of roots in terms of 'a'. We use a special formula that connects the difference, sum, and product of roots. By plugging in the given difference and calculated product, we can solve for 'a'.

🎯 Exam Tip: When given the difference of roots, remember to use the identity \( (\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta \) to connect it to the sum and product of roots.

Question 5. If one root of the equation \( 2y^2 – ay + 64 = 0 \) is twice the other then find the values of a.
Answer:
Let the roots of the equation \( 2y^2 – ay + 64 = 0 \) be \( \alpha \) and \( 2\alpha \).
For this equation, we have \( A=2, B=-a, C=64 \).
Sum of the roots: \( \alpha + 2\alpha = -\frac{B}{A} \)
\( \implies 3\alpha = -\frac{-a}{2} \)
\( \implies 3\alpha = \frac{a}{2} \) .....(1) This establishes a relationship between `\(\alpha\)` and `a`.
Product of the roots: \( \alpha \times 2\alpha = \frac{C}{A} \)
\( \implies 2\alpha^2 = \frac{64}{2} \)
\( \implies 2\alpha^2 = 32 \)
\( \implies \alpha^2 = \frac{32}{2} \)
\( \implies \alpha^2 = 16 \)
\( \implies \alpha = \pm\sqrt{16} \)
\( \implies \alpha = \pm 4 \)
Now, substitute the values of \( \alpha \) into equation (1):
Case 1: When \( \alpha = 4 \)
\( 3(4) = \frac{a}{2} \)
\( \implies 12 = \frac{a}{2} \)
\( \implies a = 12 \times 2 = 24 \)
Case 2: When \( \alpha = -4 \)
\( 3(-4) = \frac{a}{2} \)
\( \implies -12 = \frac{a}{2} \)
\( \implies a = -12 \times 2 = -24 \)
The values of a are 24 or -24. These are the possible values for 'a'.
In simple words: We name the two roots as `\(\alpha\)` and `\(2\alpha\)` since one is twice the other. Then, we use the formulas for the sum and product of roots. This lets us find `\(\alpha\)` first, and then use `\(\alpha\)` to find the values of 'a'.

🎯 Exam Tip: When roots are related (e.g., one is double the other, or one is the reciprocal of the other), represent them using a single variable (like `\(\alpha\)` and `\(2\alpha\)`) to simplify solving the problem.

Question 6. If one root of the equation \( 3x^2 + kx + 81 = 0 \) (having real roots) is the square of the other then find k.
Answer:
Let the roots of the equation \( 3x^2 + kx + 81 = 0 \) be \( \alpha \) and \( \alpha^2 \).
For this equation, we have \( A=3, B=k, C=81 \).
Sum of the roots: \( \alpha + \alpha^2 = -\frac{B}{A} \)
\( \implies \alpha + \alpha^2 = -\frac{k}{3} \) .....(1) This links 'k' with 'alpha'.
Product of the roots: \( \alpha \times \alpha^2 = \frac{C}{A} \)
\( \implies \alpha^3 = \frac{81}{3} \)
\( \implies \alpha^3 = 27 \)
To find \( \alpha \), we take the cube root of both sides:
\( \implies \alpha = \sqrt[3]{27} \)
\( \implies \alpha = 3 \) Since it says real roots, we take the real cube root.
Now, substitute the value of \( \alpha = 3 \) into equation (1):
\( 3 + (3)^2 = -\frac{k}{3} \)
\( \implies 3 + 9 = -\frac{k}{3} \)
\( \implies 12 = -\frac{k}{3} \)
\( \implies k = 12 \times (-3) \)
\( \implies k = -36 \) This is the value of 'k'.
In simple words: We name the roots `\(\alpha\)` and `\(\alpha^2\)` because one is the square of the other. We then use the formulas for the sum and product of roots. From the product formula, we can easily find the value of `\(\alpha\)`. Once we have `\(\alpha\)`, we plug it into the sum formula to find 'k'.

🎯 Exam Tip: When solving for 'k' or 'a' in root-related problems, always ensure you correctly use the sum and product of roots formulas, and substitute the relationship between the roots correctly.