Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 5 Second Degree Equations Questions and Answers Notes Pdf to clear their doubts.
SSLC Maths Chapter 5 Second Degree Equations Questions and Answers
Second Degree Equations Class 10 Questions and Answers Kerala State Syllabus
SCERT Class 10 Maths Chapter 5 Second Degree Equations Solutions
Class 10 Maths Chapter 5 Kerala Syllabus – Square Problems
(Textbook Page No.85)
Question 1.
When each side of a square was reduced by 2 metres to make a smaller square, its area became 49 square metres. What was the length of a side of the original square?
Answer:
Side of small square = \(\sqrt{49}\) = 7 m
Length of the side of original square is 9 m
Question 2.
There is a 2 metre wide path around a square ground. The area of the ground and path together is 1225 square metres. What is the area of the ground alone?
Answer:
Let x be the side
(x + 4)2 = 1225
⇒ x + 4 = 35,
x = 31 m Area of the ground alone x2 = 312 = 961 m2
(Textbook Page No.87)
Question 1.
1 added to the product of two consecutive even numbers gave 289. What are the numbers?
Answer:
If 1 is not added the product will be 288
Let x – 1 and x + 1 be the two consecutive even numbers.
(x – 1)(x + 1) = 288
⇒ x2 – 1 = 288,
x2 = 289,
x = 17 m
Numbers are 16,18
Question 2.
9 added to the product of two consecutive multiples of 6 gave 729. What are the numbers9
Answer:
x – 3 and x + 3 are two consecutive multiples of 6
If 9 is not added to the product (x – 3)(x + 3) = 720
x2 – 32 = 720,
x2 = 729,
x = 27
Numbers are 24 and 30
Question 3.
The first few terms of the arithmetic sequence 9, 11, 13, … were added and then 16 added to the sum, to get 256. How many terms were added?
Answer:
nth term of the sequence is 2n + 1
2(1 + 2 + 3 + …….. + n) + 7n = 240,
2(\(\frac{n}{2}\)(n + 1)) + 7n + 16 = 256
This can be simplified as n2 + 8n + 16 = 256
(n + 4)2= 162
⇒ n + 4 = 16,
n = 12
Sum of the first 12 terms and 16 is added we get 256
Class 10 Maths Kerala Syllabus Chapter 5 Solutions – Square Completion
(Textbook Page No.92)
Question 1.
An isosceles triangle like the one shown below must be made:
The height must be 2 metres less than the base and the area should be 12 square metres. What should be the lengths of the sides of the triangle?
Answer:
Base x , height x – 2
\(\frac{1}{2}\) × x(x – 2) – 12
⇒ x(x – 2) = 24
x2 – 2x = 24, x2 – 2x + 1 =25
(x- 1)2 = 25, x – 1 = 5, x = 6,
Height = 6 – 2 = 4
Altitude , half of the base and other side of the triangle makes a right triangle.
Side of the triangle = 5 metre.
Sides are 5 m, 5 m and 6 m
Question 2.
One side of a right triangle is one centimetre shorter than twice the side perpendicular to it and the hypotenuse is one centimetre longer than twice the length of this side. What are the lengths of the sides?
Answer:
The shorter side is x . The side perpendicular to it is 2x – 1.
Hypotenuse = 2x + 1
(2x + 1)2 = (2x- 1)2 + x2
4x2 + 4x + 1 = 4x2 – 4x + 1 + x2
x2 – 8x = 0
⇒ x = 8
Sides are 8 cm, 15 cm, 17 cm
Question 3.
A pole 2.6 metres long leans against a wall. The foot of the pole is 1 metre away from the wall. When the foot of the pole was pushed a little away, the top end slid down by the same length. How much was the foot moved?
Answer:
AC2 + AB2 = BC2
⇒ AC2 + 12 = 2.62
AC = 2.62 – 12 = 5.76,
AC = 2.4
The foot and top of the pole slid the distance
x (2.4 – x)2 + (1 + x)2 = (2.62)2
⇒ x = 1.4m
Question 4.
The product of two consecutive odd numbers is 195. What are the numbers?
Answer:
Odd numbers are x, x + 2
x(x + 2) = 195, x2 + 2x + 1 = 196
⇒ (x + 1)2 = 142, x + 1 = 14, x = 13
Numbers are 13, 15
Question 5.
How many terms of the arithmetic sequence 5, 7, 9, … must be added to get 140 as the sum ?
Answer:
Algebraic form of the sequence is 2n + 3
Sum of the first n terms = (x1 + xn) × \(\frac{n}{2}\)
(5 + 2n + 3) × \(\frac{n}{2}\) = n2 + 4n
n2 + 4n = 140
⇒ n2 + 4n + 4 = 144,
(n + 2)2 = 144,
n + 2 = 12, n = 10
Sum of first 10 terms is 140.
SCERT Class 10 Maths Chapter 5 Solutions – Two Solutions
(Textbook Page No. 96)
Question 1.
The product of a number and 2 added to it is 168. What are the numbers?
Answer:
Let x be the number.
x(x + 2) = 168,
x2 + 2x = 168
x2 + 2x + 1 = 169
⇒ (x + 1)2 = 169 x + 1 = 13, or – 13, x = 12 or – 14
Question 2.
Find two numbers with sum 4 and product 2
Answer:
Let the numbers be 2 – x and 2 + x. The sum is 4
(2 – x)(2 + x) = 2
⇒ 22 – x2 = 2,
x2 = 2, x = √2 or -√2
Numbers are 2 + √2, 2 – √2
Question 3.
Consider the arithmetic sequence 55, 45, 35, ……
(i) How many terms of this, starting from the first, must be added to get 175 as the sum?
(ii) How many terms of this, starting from the first, must be added to get 180 as the sum?
(iii) How many terms of this, starting from the first, must be added to get 0 as the sum?
Answer:
xn = -10n + 65
(i)Number of terms added is an odd number.
\(\left(\frac{n+1}{2}\right)^{t h}\)
term is the middle term. It is,
55 + (\(\frac{n+1}{2}\) – 1)d = 55 + \(\frac{n+1}{2}\) × – 10
= 55 – 5(n – 1)
= 60 – 5n
Then, 60 – 5n = \(\frac{175}{n}\)
60n – 5n2 =175
5n2 – 60n = -175
n2 – 12n = -35
n2 – 12n + 36 = 1
(n – 6)2 = 1
n – 6 = 1
n = 7
So 7 terms must be added to get sum 175.
(ii) 1th term = (-10 × 7) + 65 = -70 + 65 = -5
Sum of six terms = 175 – (-5) =180
So 6 terms must be added to get sum 180.
(iii) To get the sum 0,
Middle term must be zero,
60 – 5n = 0
5n = 60
n = 12
So 12 terms must be added to get sum 0.
Second Degree Equations Class 10 Notes Pdf
Class 10 Maths Chapter 5 Second Degree Equations Notes Kerala Syllabus
Introduction
The unit is introduced non algebraic approach of solving problems related to squaring of numbers. Three types of questions are discussed in the unit. First one is the problems which involves square directly. For example, square of a number is 1. The numbers are 1 and -1.
This can be written as x2 = 1 and solutions of this equation are 1 and -1.
- Secondly we discuss the problems whose algebraic form can be written in the square for by doing some basic operations on it. For example, the sum of the area and perimeter of a square is 140 cm2. If x is the side then we can write x2 + 4x = 140
- We add square the half of the coefficient of x on both sides. Now we get x2 + 4x + 4 = 140 + 4. This makes the equation (x + 4)2 = 122
- Third part of the unit is the problems involving two solutions. While discussing the section we recall the polynomial pictures for the better understanding of the situation.
→ The algebraic approach on problem solving is essential when the simple logic and mental calculation become difficult.
→ Square problems are the beginning in the study of second degree equations.For example, area of a square is 100 cm2. What is the length of the side of the square?
By simple reasoning, the side is 10 cm
If the square is zoomed by increasing its side by a small amount area of the new square becomes 121 cm2. How much the side increases?
By a simple logic we can see the increment is 1 cm
Algebraically, let it be x. (10 + x)2 = 121 , 10 + x = 11, x = 1cm
→ Some problems arise in slightly different way. For example : Sum of the area and perimeter of a square is 140 cm2. If x is the side then we can make an equation as x2 + 4x = 140
Adding 4 on both sides, the equation becomes x2 + 4x + 4 = 144 and it can be written as (x + 2)2 = 144, x + 2 = 12, x = 10 cm. This process is generally known as completing square.
Generally a second degree equation will have two solutions. In certain situations both solutions are acceptable.
For example, A boy’s age after 15 years will be square of his age 15 years ago. What is his present age?
This problem can be stated algebraically as (x – 15)2 = x + 15 where x is the present age. On solving this problem algebraically or otherwise we get two solutions. These are x = 10, 21.
x = 10 cannot be acceptable. So answer is 21. That is present age is 21 years
Two solutions of a second degree equation is illustrated graphically by using the graph of a second degree polynomial p(x) = x2 – 4x + 3
As we discussed in 9th standard ‘graphs of polynomials’ the graph cut horizontal line at x = 1 and x = 3. The length of verical line which cut the graph from a point (x) on the horizontal line will be the value of p(x) at x.
Here p(1) and p(3) are 0. That is x2 – 4x + 3 = 0 has solutions x = 1 and x = 3
→ A second degree equation is of the form, ax2 + bx + c = 0, where a ≠ 0, and a, b. c are real numbers.
→ Solution of a Quadratic Equation
- The solution is a value of x that satisfies the equation.
- A quadratic equation generally has two solutions.
→ Square Completion Method
- A method to solve quadratics by expressing in the form, (x + p)2 = q
- Useful in deriving the quadratic formula and solving geometrical problems.
→ Applications:
- Geometry (area and perimeter)
- Age problems
- Motion and speed
- Real-life word problems