Samacheer Kalvi 10th Maths Guide Chapter 3 Algebra Ex 3.11

Students can download Maths Chapter 3 Algebra Ex 3.11 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 3 Algebra Ex 3.11

Question 1.
Solve the following quadratic equations by completing the square method
(i) 9x² – 12x + 4 = 0
Answer:
9x² – 12x + 4 = 0
x² – \(\frac { 12x }{ 9 } \) + \(\frac { 4 }{ 9 } \) = 0 (Divided by 9)
x² – \(\frac { 4x }{ 3 } \) = \(\frac { -4 }{ 9 } \)
Add [\(\frac { 1 }{ 2 } \) (\(\frac { 4 }{ 3 } \))]² on both sides

The solution is \(\frac { 2 }{ 3 } \)

(ii) \(\frac { 5x+7 }{ x-1 } \) = 3x + 2
Answer:
(3x + 2) (x – 1) = 5x + 7
3x² – 3x + 2x – 2 = 5x + 7 ⇒ 3x² – x – 5x – 2 – 7 = 0
3x² – 6x – 9 = 0 ⇒ x² – 2x – 3 = 0 (divided by 3)
x² – 2x = 3
Adding (\(\frac { 1 }{ 2 } \) × 2)² on both sides
x² – 2x + 1 = 3 + 1
(x – 1)² = 4 ⇒ x – 1 = \(\sqrt { 4 }\)
x – 1 = ±2
x – 1 = 2 or x – 1 = -2
x = 3 or x = -1
The solution set is -1 and 3

Question 2.
Solve the following quadratic equations by formula method
(i) 2x² – 5x + 2 = 0
Answer:
a = 2, b = -5, c = 2

The solution set is \(\frac { 1 }{ 2 } \) and 2

(ii) \(\sqrt { 2 }\) f² – 6 f + 3 \(\sqrt { 2 }\) = 0
Answer:
Here a = \(\sqrt { 2 }\), b = -6 and c = 3\(\sqrt { 2 }\)

The solution set is \(\frac{3+\sqrt{3}}{\sqrt{2}}\) and \(\frac{3-\sqrt{3}}{\sqrt{2}}\)

(iii) 3y² – 20y – 23 = 0
Answer:
a = 3, b = -20, c = -23

The solution set is -1 and \(\frac { 23 }{ 3 } \)

(iv) 36y² – 12ay + (a² – b²) = 0
Answer:
Here a = 36, b = -12a, c = a² – b²

The solution set is \(\frac { (a+b) }{ 6 } \) and \(\frac { (a-b) }{ 6 } \)

Question 3.
A ball rolls down a slope and travels a distance d = t² – 0.75t feet in t seconds. Find the time when the distance travelled by the ball is 11.25 feet.
Answer:
Distance = t² – 0.75t
11.25 = t² – 0.75t
Multiply by 100
1125 = 100t² – 75t
100t² – 75t – 1125 = 0 (Divided by 25)
4t² – 3t – 45 = 0
a = 4,
b = -3,
c = -45

(time will not be negative)
The required time = \(\frac { 15 }{ 4 } \) seconds
= 3 \(\frac { 3 }{ 4 } \) second or 3.75 seconds