Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.10

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Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.10

Question 1.
Solve the following quadratic equations by factorization method
(i) 4x2 – 7x – 2 = 0
(ii) 3(p2 – 6) = p(p + 5)
(iii) \(\sqrt{a(a-7)}=3 \sqrt{2}\)
(iv) \(\sqrt{2} x^{2}+7 x+5 \sqrt{2}=0\)
(v) 2x2 – x + \(\frac{1}{8}\) = 0
Solution:
Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.10 1

(ii) 3(p2 – 6) = p(p + 5)
3p2 – 18 = p2 + 5p ⇒ 392 – 5p – 18 = 0
⇒ 2p2 – 5p – 18 = 0
⇒ (2p – 9) (p + 2) = 0 ⇒ p = \(\frac{9}{2}\), -2

(iii) \(\sqrt{a(a-7)}=3 \sqrt{2}\)
Squaring on both sides
a(a – 7) = 9 × 2
a2 – 7a – 18 = 0
a2 – 9a + 2a – 18 = 0
a(a – 9) + 2(a – 9) = 0
(a – 9) (a + 2) = 0
⇒ a = 9, a = -2

Samacheer Kalvi 10th Maths Chapter 3 Algebra Ex 3.10 2

(v) 2x2 – x + \(\frac{1}{8}\) = 0
16x2 – 8x + 1 = 0
16x2 – 4x – 4x + 1 = 0
4x(4x – 1) – 1(4x – 1) = 0
(4x – 1) (4x – 1) = 0
⇒ x = \(\frac{1}{4}, \frac{1}{4}\)

Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.10

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:
Number of games = 15
G(n) = \(\frac{n^{2}-n}{2}\)
\(\frac{n^{2}-n}{2}\) = 15
n2 – n = 30 ⇒ n2 – n – 30 = 0
⇒ n2– 6n – 5n – 30 = 0
(n – 6) (n + 5) = 0
n – 6 = 0 or n + 5 = 0
[Note: – 5 is neglected because number of team is not negative]
n = 6 or n = -5
∴ Number of teams = 6

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