Q1. Find the roots of the following quadratic equations by factorization Method:
(I) x² -3x-10 =0
by factorization method:
⇒ x² – 3x – 10 = 0
⇒ x² – ( 5 – 2 )x – 10 = 0
⇒ x² – 5x + 2x – 10 = 0
⇒ x( x – 5 ) + 2(x – 5) = 0
⇒ (x – 5)(x + 2) = 0
⇒ x – 5 = 0 ; x + 2 = 0
⇒ x = 5 ; x = – 2
⇒ x = 5,-2
(II) 2x² + x – 6 = 0
By factorization method:
⇒ 2x² + x – 6 = 0
⇒ 2x² + ( 4 – 3 )x – 6 = 0
⇒ 2x² + 4x – 3x – 6 = 0
⇒ 2x(x + 2) – 3(x – 2) = 0
⇒ (x + 2)(2x – 3) = 0
⇒ x + 2 = 0 ; 2x – 3 = 0
⇒ x = – 2 ; 2x = 3
⇒ x = – 2 ; x = 3/2
x = -2,3/2
(III) √2x²+7x+5√2 =0
By factorization method:
⇒ √2x² + (5 + 2)x + 5√2 = 0
⇒ √2x² + 5x + 2x + 5√2 = 0
⇒ x(√2x + 5) + √2(√2x + 5) = 0
⇒ (√2x + 5)(x + √2) = 0
⇒ √2x + 5 = 0 ; x + √2 = 0
⇒ √2x = -5 ; x = -√2
⇒ x = -5/√2
⇒ x = -5/√2,-√2
(IV) 2x²-x+1/8=0
⇒ 2x² – x + 1/8 = 0
⇒ 16x² – 8x + 1/8 = 0
⇒ 16x² – 8x + 1 =0×8
⇒ 16x² – (4 + 4)x + 1 = 0
⇒ 16x²- 4x – 4x + 1 = 0
⇒4x(4x – 1) – 1(4x – 1) = 0
⇒(4x – 1)(4x – 1) = 0
⇒ 4x – 1 = 0 ; 4x – 1 = 0
⇒ 4x = 1 ; 4x = 1
⇒ x = 1/4 ; x = 1/4
⇒ x=1/4,1/4
(V) 100x²-20x+1=0
⇒ 100x² – 20x + 1 = 0
⇒ 100x² – ( 10 + 10 )x + 1 = 0
⇒ 100x² – 10x – 10x + 1 = 0
⇒ 10x (10x – 1) – 1(10x – 1) = 0
⇒ (10x – 1)(10x – 1) = 0
⇒ 10x – 1 = 0 ; 10x – 1 = 0
⇒ 10x = 1 ; 10x = 1
⇒ x = 1/10 ; x = 1/10
⇒ x = 1/10,1/10
Q3. Find two numbers whose sum is 27 and whose product is 182.
Solution:
Let 1st number be = x
2nd one be = (27 – x)
x(27 – x) = 182
⇒ 27x – x² = 182
⇒ – x² + 27x = 182 = 0
⇒ – (x² – 27x + 182) = 0
⇒ x² – 27x + 182 = 0
By Factorisation method:
⇒ x² – (14 + 13)x + 182 = 0
⇒ x² – 14x – 13x + 182 = 0
⇒x(x – 14) – 13(x – 14) = 0
⇒ (x – 14)(x – 13) = 0
⇒ x – 14 = 0 ; x – 13 = 0
⇒ x = 14 ; x = 13
⇒ x = 14,13
Q4. Find two consecutive positive integers, a sum of whose square365.
Solution:
Let two consecutive positive integers are x , x+1
According to Question
(x)² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x + 1 – 365 = 0
⇒ 2x² + 2x – 364 = 0
⇒ 2(x² + x – 182) = 0
⇒ x² + x – 182 = 0/2
⇒ x² + x – 182 = 0
By Factorization method:
⇒ x² + (14 – 13)x – 182 = 0
⇒ x² + 14x – 13x – 182 = 0
⇒ x(x + 14) – 13(x + 14) = 0
⇒ (x + 14)(x – 13) = 0
x + 14 = 0 ; x – 13 = 0
⇒ x = -14 ; x = 13
x = -14, 13
Hence two consecutive positive integers 13,(13+1) = 13, 14
Q. 5 The altitude of a right triangle is 7cm less than its base. If the by hypotenuse is 13cm, find the other two sides.
Solution:
Let the base of the right triangle be = x cm.
Altitude = (x-7)cm.
Hypotenuse = 13cm.
by P.G.T
(AB)² + (BC)² = (AC)²
p²+b² = H²
⇒ (x – 7)² + (x)² = (13)²
⇒ x² – 14x + 49 + x² = 169
⇒ x² – 14x + 49 + x² – 169 = 0
⇒ 2x² – 14x – 120 = 0
⇒ 2(x² – 7x – 60) = 0
⇒ x² – 7x – 60 = 0/2
⇒ x² – 7x – 60 = 0
by Factorization method:
⇒ x² – 7x – 60 = 0
⇒ x² – (12 – 5)x – 60 = 0
⇒ x² – 12x + 5x – 60 = 0
⇒ x(x – 12) + 5(x – 12) = 0
⇒ (x – 12)(x + 5) = 0
⇒ x – 12 = 0 ; x + 5 = 0
⇒ x = 12 ; x = -5(not possible)
Hence base is = 12cm.
Altitude is =12-7 = 5cm.
Q6. 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 90 rupees, Find the number of articles produced and the cost of each article.
Solution:
Let the number of articles produced on that day = x
Now the cost of each article on that day = 3 more than twice the no of article
⇒ The cost of each article on that day = 2x + 3
The total cost of that day = 90 rupees
Number of articles × cost of each article = Total cost
⇒ x(2x + 3) = 90
⇒ 2x² + 3x = 90
⇒ 2x² + 3x – 90 = 0
By middle-term Splitting method:
⇒ 2x² + (15 – 12)x – 90 = 0
⇒ 2x² + 15x – 12x – 90 = 0
⇒x(2x + 15) – 6(2x + 15) = 0
⇒ (2x + 15)(x – 6) = 0
⇒ 2x + 15 = 0 ; x – 6 = 0
2x = -15 ; x = 6
x = -15/2 ; x = 6
Hence the number of articles produced on that day = 6
The cost of each article = 2x + 3
⇒ The cost of each article = 2×6 + 3
⇒ The cost of each article =12 + 3
⇒ The cost of each article = 15 rupees