Quadratic Equation
Mathematics Class Tenth
NCERT Exercise 4.2
Solution of a Quadratic Equation by Factorisation
Question (1) Find the roots of the following quadratic Equations by factorisation
Question (1)(i) x2 – 3x – 10 = 0
Solution:
Given, x2 – 3x – 10 = 0
By expanding middle term. Here – 3x can be written as – 5x + 2x
⇒ x2 – 5x + 2x – 10 = 0
By taking x and 2 as common, we get
⇒ x (x – 5) + 2 (x – 5) = 0
By taking (x – 5) as common, we get
⇒ (x – 5)(x + 2) = 0
Now, case – I: x – 5 = 0
⇒ x = 5
Case – II: x + 2 = 0
⇒ x = – 2
Thus, the given quadratic equation has two roots, which are – 2 and 5 Answer
Question(1) (ii)2x2 + x – 6 = 0
Solution:
Given, 2x2 + x – 6 = 0
By expanding middle term. Here x can be written as 4x – 3x
⇒ 2x2 + 4x – 3x – 6 = 0
by taking 2x as common from the first two terms and 3 as common from the last two terms
⇒ 2x(x + 2) – 3(x – 2) = 0
By taking (x + 2) as common
(x + 2)(2x – 3) = 0
Thus, Case – I:
If x + 2 = 0
∴ x = – 2
And Case – II:
If 2x – 3 = 0
∴ 2x = 3
⇒ x = 3/2
Hence, roots of the given quadratic equation are – 2 and 3/2 Answer
Question (1) (iii)√2x2 + 7x + 5 √ 2 = 0
Solution:
Given, √2x2 + 7x + 5 √2 = 0
By expanding middle term. Here 7x can be written as 2x+5x`
⇒ √ 2 x2 + 2x + 5x +5 √ 2 = 0
By taking √2from the first two terms and 5 from the last two terms as common, we get
√ 2x (x + √2) + 5x (x + √ 2) = 0
By taking (x + √2) as common, we get
⇒ (x + √ 2) (√ 2x + 5x) = 0
Now, the case – I:
If (x + √ 2) = 0
∴ x = – √2
And, the case – II:
If √2x + 5 = 0
∴√ 2x = – 5
⇒ x = – 5/√ 2
Thus, roots of given quadratic equation are – 5/√ 2 and – √ 2 Answer
Question (1)(iv) 2x2 – x + 1/8 = 0
Solution:
Given, 2x2 – x + 1/8 = 0
⇒ 16x2 – 8x + 1/8 = 0
⇒ 16x2 – 8x + 1 = 0
Now, middle term – 8x can be written as – 4x – 4x
⇒ 16x2 – 4x – 4x + 1 = 0
Now by taking 4x from the first two terms and – 1 from the last two terms as common, we get
4x (4x – 1) – 1(4x – 1) = 0
By taking 4x – 1 as common, we get
(4x – 1)(4x – 1)= 0
Here, only one case exists
⇒ 4x – 1 = 0
⇒ 4x = 1
⇒ x = 1/4
Thus, this quadratic equation has only one root, which is 1/4 Answer
Question(1)(v) 100x2 – 20x + 1 = 0
Solution:
Given, 100x2 – 20x + 1 = 0
Here middle term – 20x can be written as ( – 10x – 10x)
⇒ 100x2 – 10x – 10x + 1 = 0
Now by taking 10x from first two terms and ( – 1) from last two terms common, we get
10x (10x – 1) – 1(10x – 1) = 0
Now, by taking (10x – 1) as common we get
(10x – 1)(10x – 1) = 0
Here, only one case exist, i.e. 10x – 1 = 0
Thus, if 10x – 1 = 0
∴ 10x = 1
⇒ x = 1/10
Thus, this quadratic equation has only one root, which is 1/10 Answer
Question(2) Solve the problems given in example 1
Question (1)(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.
Solution:
Given, Number of marbles John and Jivanti together have = 45
Thus, after losing 5 marbles each of them, the number of total marbles left = 45 – (5 × 2) = 35
Now, let after losing the total number of marbles John has = x
∴ After losing number of marbles Jivanti have = 35 – x
Now, given, after losing, the product of marbles both of them have = 124
∴ x (35 – x) = 124
⇒ 35x – x2 = 124
⇒ – x2 + 35x – 124 = 0
⇒ – x2 + 31x + 4x – 124 = 0
[∵ middle term (+ 35x) can be written as (31x + 4x)]
Now, by taking ( – x) from the first two terms and (4) from the last two terms common, we get
– x(x – 31) + 4(x – 31) = 0
Now, by taking (x – 31) as common, we get
(x – 31)( – x + 4) = 0
Now, the Case – I :
If, (x – 31) = 0
:. x = 31
And the Case – II :
If – x + 4 = 0
⇒ – x = 4 ⇒ x = – 4
Since, in case II, the number of marbles is negative ( – ), which is not possible.
Thus, only case – I is taking into consideration
Thus, No. of marbles John have = x + 5 = 31 + 5 = 36
Since, together they had total of 45 marbles, thus, number of marbles Jivanti have = 45 – 36 = 9
∴ John's marbles = 36, And Jivanti's marbles = 9 Answer
Question(2)(ii) A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs 750. We would like to find out the number of toys produced in on that day.
Solution:
Let number of toys produced in a day = x
∴ cost of production of each toy = 55 – x
Given, On a particular day total cost of production = Rs 750
∴ number of toys xx cost of production = total cost of production
⇒ x × (55 – x) = 750
⇒ 55x – x2 = 750
⇒ – x2 + 55x – 750 = 0
⇒ x2 – 55x + 750 = 0
[By taking minus as common]
After expanding the above equation, we get
x2 – 25x – 30x + 750 = 0
⇒ x(x – 25) – 30(x – 25) = 0
⇒ (x – 25)(x – 30) = 0
Now, if (x – 25) = 0
∴ x = 25
And if (x – 30) = 0
⇒ x = 30
Thus, the number of toys = 25 or 30 Answer
Question (3) Find two numbers whose sum is 27 and product is 182
Solution:
Let first number is x
Since the sum of two numbers is equal to 27
∴ second number = 27 – x
Now, as given, the product of given numbers = 182
∴ x(27 – x) = 182
⇒ 27x – x = 182
⇒ – x2 + 27x = 182
⇒ – x2 + 27x – 182 = 0
After getting minus as common, we get
x2 – 27x + 182 = 0
Now, after expanding the middle term, we get
x2 – 14x – 13x + 182 = 0
⇒ x(x – 14) – 13(x – 14) = 0
⇒ (x – 14)(x – 13) = 0
Now, if x – 14 = 0
∴ x = 14
Thus, the second number = 27 – 14 = 13
Thus, the numbers are 14 and 13 Answer
Question(4) Find two consecutive positive integers, the sum of whose squares is 365
Solution:
Let, first number = x
Therefore, second number = x + 1
Now, as given, x2 + (x + 1)2 = 365
⇒ x2 + x2 + 2x + 1 = 365
[∵ (a + b)2 = a2 + b2 + 2ab]
⇒ 2x2 + 2x + 1 – 365 = 0
⇒ 2x2 + 2x – 364 = 0
⇒ 2(x2 + x – 182) = 0
⇒ x2 + x – 182 = 0
After expanding the middle term, we get
⇒ x2 + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
Now, if x + 14 = 0
∴ x = – 14
This is not possible, because the given integers are positive
And, if x – 13 = 0
∴ x = 13
Thus, the next consecutive number = 13 + 1 = 14
Thus, the numbers are 13 and 14 Answer
Question: (5) The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13cm, find the other two sides
Solution:
Given, hypotenuse = 13cm
Let, base = x
∴ as given, altitude i.e. height = base – 7cm
i.e. height = x – 7 cm
We know that, hypotenuse2 = height2 + base2
(13 cm)2 = (x – 7)2 + x2
⇒ 169 = x2 + 49 – 14x + x2
[∵ (a + b)2 = a2 + b2 + 2ab]
⇒ 169 = 2x2 – 14x + 49
⇒ 2x2 – 14x + 49 – 169 = 0
⇒ 2x2 – 14x – 120 = 0
⇒ 2(x2 – 7x – 60) = 0
⇒ x2 – 7x – 60 = 0
After expanding the middle term, we get
⇒ x2 – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12)= 0
⇒ (x – 12)(x + 5) = 0
Now, if x + 5 = 0
⇒ x = – 5
This is not possible
And, if x – 12 = 0
∴ x = 12 = base
∵ height = base – 7
Therefore, height = 12 – 7 = 5
Thus, the height = 5 and base = 12
Thus, the other two sides are 5 and 12 Answer
Question (6) A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs 90, find the number of articles produced and the cost of each article.
Solution:
Let total number of articles produced on the given day = x
Therefore, cost of one article as given, = 2x + 3
As given, total cost of production on that day = Rs 90
∴ x(2x + 3) = 90
⇒ 2x2 + 3x = 90
⇒ 2x2 + 3x – 90 = 0
After expanding the middle term, we get
2x2 – 12x + 15x – 90 = 0
⇒ 2x(x – 6) + 15(x – 6) = 0
⇒ (2x + 15)(x – 6) = 0
Now, if (2x + 15) = 0
∴ 2x = – 15
⇒ x = – 15/2
This is not possible as it is in negative
And, if x – 6 = 0
∴ x = 6
Thus, number of articles produced on the given day = 6
Now, ∵ cost of one article = 2x + 3
= 2 × 6 + 3 = 15
Thus, number of articles = 6
And cost of each article = Rs 15 Answer
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