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Quadratic equation Class 10^th

Que. 1. Find the roots of the following quadratic equations by factorization method:

1. x² – 3x – 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

2. 2x² + x – 6 = 0

 2x² + 4x – 3x – 6 = 0

2x (x + 2) – 3 (x + 2) = 0

(x + 2)  (2x – 3)

X + 2 = 0            2x – 3 = 0

X = -2                 2x = 3

16x² – 8x + 1 = 0 x 8

16x² – 8x + 1 = 0

16x² – 4x – 4x + 1 = 0

4x (4x – 1) – 1 (4x – 1) = 0

(4x – 1)  (4x – 1)

4x – 1 = 0               4x – 1 = 0

4x = 1                     4x = 1

5. 100x² – 20x + 1 = 0

100x² – 10x – 10x + 1 = 0

10x (10x – 1) – 1 (10x – 1) = 0

(10x – 1)  (10x – 1)

10x – 1 = 0             10x – 1 = 0

10x = 1                   10x = 1

Que. 2. Solve the problems given in Example 1.

1. Let John have marbles – x

Let Javanti have marbles – 45 – x

As per the question,

Both lose five marbles –

John has marbles = x – 5

Javanti has marbles = 45 – x -5

                            = 40 – x 

(x – 5)  (40 – x) = 124

40x – x² – 200 + 5x = 124

-x² + 45x – 324 = 0

X² – 45x + 324 = 0
X² – 36x – 9x + 324 = 0

X (x – 36) – 9 (x – 36) = 0

(x – 36)  (x – 9) = 0

X – 36 = 0            x – 9 = 0

X = 36                  x = 9

2. Let the number of toys manufactured in a day = x

Cost of one toy = (55 – x) rupees

 X (55 – x) = 750

55x – x² = 750

55x – x² -750 = 0

X² -30x -25x +750 = 0

X (x -30) -25 (x -30) = 0

(x -30) (x -25) = 0

x -30 = 0                      x -25 = 0

x = 30                           x = 25

Que. 3. Find two numbers whose sum is 27 and their product is 182.

Let the first number = x

Let the second number = 27 – x

As per the question,

X x (27 – x) = 182

27x – x² = 182

X² – 27x + 182 = 0

X² – 13x – 14x + 182 = 0

X (x – 13) – 14 (x – 13) = 0

(x – 13)  (x – 14) = 0

X = 13 = 0           x – 14 = 0

X = 13                  x =  14

Que. 4. Find two consecutive positive integers whose squares sum to 365.

Let the first positive integer be = x

Let the second positive integer be = x + 1

( 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

X² + 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

-14 is a negative value, so the first positive integer is 13 and the second positive integer is 14.

Que. 5. The height of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Let base of right angled triangle = x

Let height of right angled triangle = (x – 7)

Let hypotenuse of right angled triangle = 13 cm

From Pythagoras theorem,

(hypotenuse)²  =  (base)²  +  (perpendicular)²

(AC)²   =   (BC)²  +  (AB)²

(13)²    =   (x)² + (x – 7)²

169     =    x² + x² – 14x + (7)²

169 = 2x² – 14x + 49

2x² – 14x + 49 – 169 = 0

2x² – 14x – 120 = 0

2 (x² – 7x – 60) = 0

X² – 7x – 60 = 0/2 = 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

                          Cancelling,

Base = x = 12 cm

Height = x – 7

= 12 – 7

= 5 cm

Que. 6. A cottage industry produces some number of pots per day. On one particular day, it was observed that the manufacturing cost (in ₹) of each piece was 3 times more than twice the number of pots manufactured that day. If the total manufacturing cost for that day was ₹90, find the number of pots manufactured and the cost of each piece.

Let the number of pots = x

Cost of a pot = 2x + 3

X (2x + 3) = 90

2x² + 3x = 90

2x² + 3x – 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

Number of pots x = 6

Cost of one pot = 2x + 3

                          = 2 x 6 + 3

                          = 15 rupees