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**Q.1. Find the roots of the following quadratic equations by factorisation:****(i) x ^{2} – 3x – 10 = 0**

Taking LHS,

=>x

=>x(x – 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x

Therefore, x – 5 = 0 or x + 2 = 0

=> x = 5 or x = -2

Taking LHS,

=> 2x

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2x

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

Taking LHS,

=> √2 x

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

Taking LHS,

=1/8 (16x

= 1/8 (16x

= 1/8 (4x(4x

= 1/8 (4x – 1)

The roots of this equation, 2x

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

Taking LHS,

= 100x

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)

The roots of this equation, 100x

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

(i) Let us say, the number of marbles John have = x.

Therefore, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124.

∴ (x – 5)(40 – x) = 124

⇒ x

⇒ x

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

Therefore,

If, John’s marbles = 36,

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9,

Then, Jivanti’s marbles = 45 – 9 = 36

(ii) Let us say, number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x

⇒ x

⇒ x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

Let us say, first number be x and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x

⇒ x

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

Let us say, the two consecutive positive integers be x and x + 1.

Therefore, as per the given questions,

x

⇒ x

⇒ 2x

⇒ x

⇒ x

⇒ x(x + 14) -13(x + 14) = 0

⇒ (x + 14)(x – 13) = 0

Thus, either, x + 14 = 0 or x – 13 = 0,

⇒ x = – 14 or x = 13

since, the integers are positive, so x can be 13, only.

∴ x + 1 = 13 + 1 = 14

Therefore, two consecutive positive integers will be 13 and 14.

Let us say, the base of the right triangle be x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base

∴ x

⇒ x

⇒ 2x

⇒ x

⇒ x

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

Let us say, the number of articles produced be x.

Therefore, cost of production of each article = Rs (2x + 3)

Given, total cost of production is Rs.90

∴ x(2x + 3) = 90

⇒ 2x

⇒ 2x

⇒ x(2x + 15) -6(2x + 15) = 0

⇒ (2x + 15)(x – 6) = 0

Thus, either 2x + 15 = 0 or x – 6 = 0

⇒ x = -15/2 or x = 6

As the number of articles produced can only be a positive integer, therefore, x can only be 6.

Hence, number of articles produced = 6

Cost of each article = 2 × 6 + 3 = Rs 15.

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