Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.10

Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.10

Question 1. Solve the following quadratic equations by factorization method
(i) \( 4x^2 – 7x − 2 = 0 \)
(ii) \( 3(p^2 – 6) = p(p + 5) \)
(iii) \( \sqrt{a(a-7)} = 3 \sqrt{2} \)
(iv) \( \sqrt { 2 } x^2 + 7x + 5\sqrt { 2 } = 0 \)
(v) \( 2x^2 -x + \frac { 1 }{ 8 } = 0 \)
Answer:
(i) To solve \( 4x^2 – 7x − 2 = 0 \) by factorization:
First, split the middle term, \( -7x \), into \( -8x + x \):
\( 4x^2 - 8x + x - 2 = 0 \)
Now, group the terms and factor out common parts:
\( 4x(x - 2) + 1(x - 2) = 0 \)
This gives a common factor of \( (x - 2) \):
\( (x - 2)(4x + 1) = 0 \)
For the product of two factors to be zero, at least one of them must be zero:
\( x - 2 = 0 \)
\( \implies x = 2 \)
Or
\( 4x + 1 = 0 \)
\( \implies 4x = -1 \)
\( \implies x = \frac{-1}{4} \)
So, the roots of the equation are \( 2 \) and \( \frac{-1}{4} \). Factorization helps find the values of 'x' that make the equation true.

(ii) To solve \( 3(p^2 – 6) = p(p + 5) \) by factorization:
First, open the brackets and rearrange the terms to form a standard quadratic equation \( ap^2 + bp + c = 0 \):
\( 3p^2 - 18 = p^2 + 5p \)
Subtract \( p^2 \) and \( 5p \) from both sides:
\( 3p^2 - p^2 - 5p - 18 = 0 \)
\( 2p^2 - 5p - 18 = 0 \)
Next, split the middle term, \( -5p \), into \( -9p + 4p \):
\( 2p^2 - 9p + 4p - 18 = 0 \)
Group the terms and factor:
\( p(2p - 9) + 2(2p - 9) = 0 \)
Factor out the common term \( (2p - 9) \):
\( (2p - 9)(p + 2) = 0 \)
Set each factor to zero to find the values of \( p \):
\( 2p - 9 = 0 \)
\( \implies 2p = 9 \)
\( \implies p = \frac{9}{2} \)
Or
\( p + 2 = 0 \)
\( \implies p = -2 \)
Therefore, the roots of the equation are \( \frac{9}{2} \) and \( -2 \). It is important to simplify the equation first before trying to factorize.

(iii) To solve \( \sqrt{a(a-7)} = 3 \sqrt{2} \):
To remove the square root, square both sides of the equation:
\( (\sqrt{a(a-7)})^2 = (3\sqrt{2})^2 \)
\( a(a - 7) = 3^2 (\sqrt{2})^2 \)
\( a^2 - 7a = 9 \times 2 \)
\( a^2 - 7a = 18 \)
Rearrange into a standard quadratic equation:
\( a^2 - 7a - 18 = 0 \)
Split the middle term, \( -7a \), into \( -9a + 2a \):
\( a^2 - 9a + 2a - 18 = 0 \)
Group the terms and factor:
\( a(a - 9) + 2(a - 9) = 0 \)
Factor out the common term \( (a - 9) \):
\( (a - 9)(a + 2) = 0 \)
Set each factor to zero:
\( a - 9 = 0 \)
\( \implies a = 9 \)
Or
\( a + 2 = 0 \)
\( \implies a = -2 \)
Thus, the roots of the equation are \( 9 \) and \( -2 \). Squaring both sides helps convert radical equations into solvable quadratic forms.

(iv) To solve \( \sqrt { 2 } x^2 + 7x + 5\sqrt { 2 } = 0 \):
We will use the factorization method by splitting the middle term. We need two numbers that multiply to \( (\sqrt{2})(5\sqrt{2}) = 10 \) and add up to \( 7 \). These numbers are \( 2 \) and \( 5 \).
\( \sqrt { 2 } x^2 + 2x + 5x + 5\sqrt { 2 } = 0 \)
Group the terms and factor out common parts. Remember that \( 2 = \sqrt{2} \times \sqrt{2} \):
\( \sqrt { 2 } x (x + \sqrt { 2 }) + 5(x + \sqrt { 2 }) = 0 \)
Factor out the common term \( (x + \sqrt { 2 }) \):
\( (x + \sqrt { 2 })(\sqrt { 2 }x + 5) = 0 \)
Set each factor to zero to find the values of \( x \):
\( x + \sqrt { 2 } = 0 \)
\( \implies x = -\sqrt { 2 } \)
Or
\( \sqrt { 2 }x + 5 = 0 \)
\( \implies \sqrt { 2 }x = -5 \)
\( \implies x = \frac{-5}{\sqrt{2}} \)
The roots are \( -\sqrt { 2 } \) and \( \frac{-5}{\sqrt{2}} \). Splitting the middle term correctly is key for factorization.

(v) To solve \( 2x^2 -x + \frac { 1 }{ 8 } = 0 \):
First, clear the fraction by multiplying the entire equation by 8:
\( 8 \times (2x^2 - x + \frac{1}{8}) = 8 \times 0 \)
\( 16x^2 - 8x + 1 = 0 \)
This is a perfect square trinomial, \( (4x - 1)^2 \). We can also factor by splitting the middle term \( -8x \) into \( -4x - 4x \):
\( 16x^2 - 4x - 4x + 1 = 0 \)
Group the terms and factor:
\( 4x(4x - 1) - 1(4x - 1) = 0 \)
Factor out the common term \( (4x - 1) \):
\( (4x - 1)(4x - 1) = 0 \)
Set each factor to zero:
\( 4x - 1 = 0 \)
\( \implies 4x = 1 \)
\( \implies x = \frac{1}{4} \)
Since both factors are the same, the equation has two identical roots.
The roots are \( \frac{1}{4} \) and \( \frac{1}{4} \). Multiplying by the denominator first makes the equation easier to solve.
In simple words: For each equation, we found the values of the variable (x, p, or a) that make the equation true. We did this by breaking the equations into simpler multiplication parts.

🎯 Exam Tip: Always make sure to set up the quadratic equation in the standard form \( ax^2 + bx + c = 0 \) before attempting to factorize.

Question 2. The number of volleyball games that must be scheduled in a league with n teams is given by \( G(n) = \frac{n^{2}-n}{2} \) where each team plays with every other team exactly once. A league schedules 15 games. How many teams are in the league?
Answer: We are given the formula for the number of games \( G(n) \) in a league with \( n \) teams:
\( G(n) = \frac{n^{2}-n}{2} \)
We are told that the league schedules 15 games, so \( G(n) = 15 \). Substitute this into the formula:
\( 15 = \frac{n^{2}-n}{2} \)
Multiply both sides by 2 to remove the fraction:
\( 30 = n^2 - n \)
Rearrange the terms to form a standard quadratic equation:
\( n^2 - n - 30 = 0 \)
Now, factorize the quadratic equation. We need two numbers that multiply to \( -30 \) and add up to \( -1 \). These numbers are \( -6 \) and \( 5 \).
\( n^2 - 6n + 5n - 30 = 0 \)
Group the terms and factor:
\( n(n - 6) + 5(n - 6) = 0 \)
Factor out the common term \( (n - 6) \):
\( (n - 6)(n + 5) = 0 \)
Set each factor to zero to find the possible values of \( n \):
\( n - 6 = 0 \)
\( \implies n = 6 \)
Or
\( n + 5 = 0 \)
\( \implies n = -5 \)
Since the number of teams cannot be negative, we discard \( n = -5 \).
Therefore, there are 6 teams in the league. The value of 'n' must be a positive whole number because it represents a count of teams.
In simple words: We used the given formula for games and the total games played to find how many teams are in the league. We solved a simple equation and found there are 6 teams.

🎯 Exam Tip: When solving word problems involving quantities like the number of teams or people, always check that your answer makes sense in the real world (e.g., quantities cannot be negative or fractional).