Given:

(i) Total marbles with John and Jivanti = $45$. After losing $5$ marbles each, the product of the remaining marbles is $124$.

(ii) Let $x$ be the number of toys produced. The cost of production per toy = $55 - x$. Total cost of production = $Rs. 750$.

To Find:

(i) The initial number of marbles John and Jivanti had.

(ii) The number of toys produced on that day.

Part (i): Solving for Marbles

Step 1: Define variables.

Let the number of marbles John had be $x$.

Since the total number of marbles is $45$, the number of marbles Jivanti had is $(45 - x)$.

Step 2: Formulate the equation based on the condition.

After losing $5$ marbles each:

John's remaining marbles = $x - 5$

Jivanti's remaining marbles = $(45 - x) - 5 = 40 - x$

The product of these is $124$:

$(x - 5)(40 - x) = 124$

Step 3: Simplify the quadratic equation.

$40x - x^2 - 200 + 5x = 124$

$-x^2 + 45x - 200 = 124$

$x^2 - 45x + 324 = 0$ [Rearranging terms to standard form $ax^2 + bx + c = 0$]

Step 4: Solve by factorization.

We need two numbers that multiply to $324$ and add to $-45$. These are $-36$ and $-9$.

$x^2 - 36x - 9x + 324 = 0$

$x(x - 36) - 9(x - 36) = 0$

$(x - 36)(x - 9) = 0$

Therefore, $x = 36$ or $x = 9$.

If John had $36$ marbles, Jivanti had $9$. If John had $9$ marbles, Jivanti had $36$.

Final Answer (i): John and Jivanti started with 36 and 9 marbles respectively.

Part (ii): Solving for Toys

Step 1: Define variables.

Let the number of toys produced be $x$.

Cost of production per toy = $(55 - x)$

Step 2: Formulate the equation.

Total cost = (Number of toys) $\times$ (Cost per toy)

$x(55 - x) = 750$

Step 3: Simplify the quadratic equation.

$55x - x^2 = 750$

$x^2 - 55x + 750 = 0$

Step 4: Solve by factorization.

We need two numbers that multiply to $750$ and add to $-55$. These are $-25$ and $-30$.

$x^2 - 30x - 25x + 750 = 0$

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

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

Therefore, $x = 30$ or $x = 25$.

Final Answer (ii): The number of toys produced on that day was either 25 or 30.

Given: A quadratic equation $100x^2 - 20x + 1 = 0$.

To find: The roots of the given quadratic equation using the method of factorisation.

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $100x^2 - 20x + 1 = 0$ with the standard form:
$a = 100$
$b = -20$
$c = 1$

Step 2: Determine the splitting of the middle term.
To factorise by splitting the middle term, we need to find two numbers such that:
1. Their product is equal to $a \times c = 100 \times 1 = 100$.
2. Their sum is equal to $b = -20$.

Let the two numbers be $p$ and $q$.
$p \times q = 100$
$p + q = -20$
Considering the factors of $100$: $(-10) \times (-10) = 100$ and $(-10) + (-10) = -20$.
Thus, the two numbers are $-10$ and $-10$.

Step 3: Rewrite the middle term and factorise by grouping.
Substitute $-20x$ with $-10x - 10x$ in the original equation:
$100x^2 - 10x - 10x + 1 = 0$

Group the terms into two pairs:
$(100x^2 - 10x) - (10x - 1) = 0$

Factor out the greatest common factor from each group:
$10x(10x - 1) - 1(10x - 1) = 0$

Step 4: Extract the common binomial factor.
$(10x - 1)(10x - 1) = 0$
This can be written as:
$(10x - 1)^2 = 0$

Step 5: Solve for $x$.
To find the roots, set the factor equal to zero:
$10x - 1 = 0$
$10x = 1$
$x = \frac{1}{10}$

Since the equation is a perfect square, both roots are identical:
$x = \frac{1}{10}, \frac{1}{10}$

Final Answer: The roots of the quadratic equation $100x^2 - 20x + 1 = 0$ are $\frac{1}{10}$ and $\frac{1}{10}$.

Given: The quadratic equation $2x^2 - x + \frac{1}{8} = 0$.

To find: The roots of the given quadratic equation by the method of factorisation.

Step 1: Simplifying the Equation
To make the factorisation process easier, we first eliminate the fraction by multiplying the entire equation by $8$.

$8 \times (2x^2 - x + \frac{1}{8}) = 8 \times 0$

$16x^2 - 8x + 1 = 0$

Step 2: Splitting the Middle Term
We need to factorise the quadratic expression $16x^2 - 8x + 1$. We look for two numbers such that:
1. Their product is equal to the product of the coefficient of $x^2$ and the constant term: $16 \times 1 = 16$.
2. Their sum is equal to the coefficient of $x$: $-8$.

The two numbers that satisfy these conditions are $-4$ and $-4$, since $(-4) \times (-4) = 16$ and $(-4) + (-4) = -8$.

Step 3: Factorising by Grouping
Rewrite the middle term $-8x$ as $-4x - 4x$:

$16x^2 - 4x - 4x + 1 = 0$

Group the terms into two pairs:

$(16x^2 - 4x) - (4x - 1) = 0$

Factor out the common terms from each group:

$4x(4x - 1) - 1(4x - 1) = 0$

Step 4: Extracting the Factors
Now, factor out the common binomial $(4x - 1)$:

$(4x - 1)(4x - 1) = 0$

$(4x - 1)^2 = 0$

Step 5: Finding the Roots
To find the roots, set each factor equal to zero [By the Zero Product Property]:

$4x - 1 = 0$

$4x = 1$

$x = \frac{1}{4}$

Since both factors are identical, the equation has two equal real roots.

Final Answer: The roots of the quadratic equation are $x = \frac{1}{4}$ and $x = \frac{1}{4}$.

Given: Two consecutive positive integers. The sum of the squares of these integers is $365$.

To Find: The two consecutive positive integers.

Step 1: Defining the Variables
Let the first positive integer be $x$.
Since the integers are consecutive, the next integer must be $x + 1$.
[Assumption: $x$ is a positive integer, so $x > 0$].

Step 2: Formulating the Equation
According to the problem statement, the sum of the squares of these two integers is $365$.
Mathematically, this is expressed as:
$x^2 + (x + 1)^2 = 365$

Step 3: Expanding and Simplifying the Equation
Apply the algebraic identity $(a + b)^2 = a^2 + 2ab + b^2$ to expand $(x + 1)^2$:
$x^2 + (x^2 + 2x + 1) = 365$
Combine like terms:
$2x^2 + 2x + 1 = 365$
Subtract $365$ from both sides to set the quadratic equation to the standard form $ax^2 + bx + c = 0$:
$2x^2 + 2x + 1 - 365 = 0$
$2x^2 + 2x - 364 = 0$

Step 4: Simplifying the Quadratic Equation
Divide the entire equation by $2$ to simplify the coefficients:
$x^2 + x - 182 = 0$

Step 5: Solving by Factorization
We need to find two numbers that multiply to $-182$ and add to $1$ (the coefficient of $x$).
The factors of $182$ are: $1 \times 182, 2 \times 91, 7 \times 26, 13 \times 14$.
Since the product is negative and the sum is positive, we choose $14$ and $-13$.
$x^2 + 14x - 13x - 182 = 0$
Group the terms:
$x(x + 14) - 13(x + 14) = 0$
Factor out the common binomial $(x + 14)$:
$(x + 14)(x - 13) = 0$

Step 6: Finding the Roots
Set each factor to zero:
1) $x + 14 = 0 \implies x = -14$
2) $x - 13 = 0 \implies x = 13$

Step 7: Evaluating the Results
The problem specifies that the integers must be positive.
Therefore, we reject $x = -14$.
We accept $x = 13$.

Step 8: Determining the Consecutive Integer
If the first integer is $x = 13$, then the second consecutive integer is:
$x + 1 = 13 + 1 = 14$.

Verification:
$13^2 + 14^2 = 169 + 196 = 365$. The condition is satisfied.

Final Answer: The two consecutive positive integers are 13 and 14.

Given: A quadratic equation $2x^2 + x - 6 = 0$.

To find: The roots of the given quadratic equation using the method of factorisation (splitting the middle term).

Step 1: Identify the coefficients of the quadratic equation.
The standard form of a quadratic equation is $ax^2 + bx + c = 0$.
Comparing $2x^2 + x - 6 = 0$ with the standard form:
$a = 2$
$b = 1$
$c = -6$

Step 2: Determine the product and the sum for splitting the middle term.
To factorise by splitting the middle term, we need to find two numbers such that:
1. Their product is equal to $a \times c = 2 \times (-6) = -12$.
2. Their sum is equal to $b = 1$.

Step 3: Find the two numbers.
We look for factors of $-12$ that add up to $1$:
Factors of $-12$: $(-1, 12), (1, -12), (-2, 6), (2, -6), (-3, 4), (3, -4)$.
Checking the sums:
$-3 + 4 = 1$.
Thus, the two numbers are $4$ and $-3$.

Step 4: Rewrite the middle term and factorise by grouping.
Rewrite the equation $2x^2 + x - 6 = 0$ as:
$2x^2 + 4x - 3x - 6 = 0$ [Splitting the middle term $x$ into $4x - 3x$]
Group the terms:
$(2x^2 + 4x) - (3x + 6) = 0$
Factor out the common terms from each group:
$2x(x + 2) - 3(x + 2) = 0$ [Factoring out $2x$ from the first group and $3$ from the second group]
$(2x - 3)(x + 2) = 0$ [Taking $(x + 2)$ as a common factor]

Step 5: Find the roots by setting each factor to zero.
According to the Zero Product Property, if the product of two factors is zero, at least one of the factors must be zero.
Case 1: $2x - 3 = 0$
$2x = 3$
$x = \frac{3}{2}$

Case 2: $x + 2 = 0$
$x = -2$

Final Answer: The roots of the quadratic equation $2x^2 + x - 6 = 0$ are $x = \frac{3}{2}$ and $x = -2$.

Given: A quadratic equation $x^2 - 3x - 10 = 0$.

To Find: The roots of the given quadratic equation using the method of factorisation.

Step 1: Understanding the Factorisation Method (Splitting the Middle Term)
To factorise a quadratic equation of the form $ax^2 + bx + c = 0$, we look for two numbers, say $p$ and $q$, such that:
1. The product of the numbers is equal to the product of the coefficient of $x^2$ and the constant term: $p \times q = a \times c$.
2. The sum of the numbers is equal to the coefficient of $x$: $p + q = b$.
In the given equation $x^2 - 3x - 10 = 0$, we have $a = 1$, $b = -3$, and $c = -10$.
Therefore, we need two numbers such that:
Product ($p \times q$) = $1 \times (-10) = -10$
Sum ($p + q$) = $-3$

Step 2: Identifying the Factors
We list the pairs of factors of $-10$:
- $(1, -10) \implies 1 + (-10) = -9$
- $(-1, 10) \implies -1 + 10 = 9$
- $(2, -5) \implies 2 + (-5) = -3$
- $(-2, 5) \implies -2 + 5 = 3$
The pair that satisfies both conditions is $2$ and $-5$.

Step 3: Splitting the Middle Term
We rewrite the middle term $-3x$ as $2x - 5x$:
$x^2 + 2x - 5x - 10 = 0$

Step 4: Grouping and Factoring
Group the terms into two pairs:
$(x^2 + 2x) - (5x + 10) = 0$
Factor out the common terms from each group:
$x(x + 2) - 5(x + 2) = 0$
Now, factor out the common binomial $(x + 2)$:
$(x + 2)(x - 5) = 0$

Step 5: Finding the Roots
According to the Zero Product Property, if the product of two factors is zero, then at least one of the factors must be zero.
Case 1: $x + 2 = 0$
$x = -2$
Case 2: $x - 5 = 0$
$x = 5$

Final Answer: The roots of the quadratic equation $x^2 - 3x - 10 = 0$ are $x = -2$ and $x = 5$.

Given:

1. The sum of two numbers is $27$.

2. The product of the same two numbers is $182$.

To Find:

The two numbers.

Step 1: Defining the Variables

Let the first number be $x$.

Since the sum of the two numbers is $27$, the second number can be expressed as $(27 - x)$.

Step 2: Formulating the Quadratic Equation

According to the problem, the product of these two numbers is $182$. Therefore, we can set up the following equation:

$x(27 - x) = 182$

Step 3: Simplifying the Equation

Distribute $x$ into the parentheses:

$27x - x^2 = 182$

Rearrange the terms to form a standard quadratic equation of the form $ax^2 + bx + c = 0$:

$-x^2 + 27x - 182 = 0$

Multiply the entire equation by $-1$ to make the leading coefficient positive:

$x^2 - 27x + 182 = 0$

Step 4: Solving by Factorization (Splitting the Middle Term)

We need to find two numbers that multiply to $182$ and add up to $-27$.

Prime factorization of $182$:

$182 = 2 \times 91 = 2 \times 7 \times 13 = 14 \times 13$

Since the sum must be $-27$, we choose $-14$ and $-13$ because $(-14) + (-13) = -27$ and $(-14) \times (-13) = 182$.

Rewrite the middle term:

$x^2 - 14x - 13x + 182 = 0$

Step 5: Grouping and Factoring

Group the terms:

$(x^2 - 14x) - (13x - 182) = 0$

Factor out the common terms from each group:

$x(x - 14) - 13(x - 14) = 0$

Factor out the common binomial $(x - 14)$:

$(x - 14)(x - 13) = 0$

Step 6: Finding the Roots

Set each factor to zero [Using the Zero Product Property]:

Case 1: $x - 14 = 0 \implies x = 14$

Case 2: $x - 13 = 0 \implies x = 13$

If the first number is $14$, the second number is $27 - 14 = 13$.

If the first number is $13$, the second number is $27 - 13 = 14$.

Final Answer: The two numbers are 13 and 14.

Given: A quadratic equation $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$.

To Find: The roots of the given quadratic equation using the method of factorisation.

Step 1: Identify the coefficients of the quadratic equation.

The standard form of a quadratic equation is $ax^2 + bx + c = 0$. Comparing our equation $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ with the standard form, we have:

$a = \sqrt{2}$

$b = 7$

$c = 5\sqrt{2}$

Step 2: Determine the product $a \times c$ and find two numbers that sum to $b$.

We need to find two numbers whose product is $a \times c$ and whose sum is $b$.

Product = $a \times c = \sqrt{2} \times 5\sqrt{2} = 5 \times (\sqrt{2})^2 = 5 \times 2 = 10$.

Sum = $b = 7$.

We look for two numbers $p$ and $q$ such that $p \times q = 10$ and $p + q = 7$.

The factors of $10$ are $(1, 10)$ and $(2, 5)$. Since $2 + 5 = 7$, the required numbers are $2$ and $5$.

Step 3: Split the middle term.

Rewrite the middle term $7x$ as $(2x + 5x)$:

$\sqrt{2}x^2 + 2x + 5x + 5\sqrt{2} = 0$

Step 4: Factor by grouping.

Group the terms to extract common factors:

$(\sqrt{2}x^2 + 2x) + (5x + 5\sqrt{2}) = 0$

Note that $2$ can be written as $\sqrt{2} \times \sqrt{2}$. Thus, $\sqrt{2}x^2 + (\sqrt{2} \times \sqrt{2})x = \sqrt{2}x(x + \sqrt{2})$.

$\sqrt{2}x(x + \sqrt{2}) + 5(x + \sqrt{2}) = 0$

Step 5: Factor out the common binomial $(x + \sqrt{2})$.

$(x + \sqrt{2})(\sqrt{2}x + 5) = 0$

Step 6: Solve for $x$ by setting each factor to zero.

[By the Zero Product Property, if $A \times B = 0$, then $A = 0$ or $B = 0$]

Case 1: $x + \sqrt{2} = 0$

$x = -\sqrt{2}$

Case 2: $\sqrt{2}x + 5 = 0$

$\sqrt{2}x = -5$

$x = -\frac{5}{\sqrt{2}}$

To rationalise the denominator: $x = -\frac{5}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = -\frac{5\sqrt{2}}{2}$

Final Answer: The roots of the quadratic equation are $-\sqrt{2}$ and $-\frac{5\sqrt{2}}{2}$.

Given:

• A right-angled triangle.
• The length of the hypotenuse is $13$ cm.
• The altitude (height) is $7$ cm less than the base.

To find:

• The lengths of the base and the altitude of the triangle.

Step 1: Defining Variables

Let the base of the right triangle be $x$ cm.

According to the problem, the altitude is $7$ cm less than the base.

Therefore, the altitude = $(x - 7)$ cm.

Step 2: Applying the Pythagorean Theorem

In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (Pythagorean Theorem):

$(\text{Hypotenuse})^2 = (\text{Base})^2 + (\text{Altitude})^2$

Substituting the given values:

$13^2 = x^2 + (x - 7)^2$

Step 3: Expanding and Simplifying the Equation

$169 = x^2 + (x^2 - 14x + 49)$ [Using the identity $(a - b)^2 = a^2 - 2ab + b^2$]

$169 = 2x^2 - 14x + 49$

Subtract $169$ from both sides to set the quadratic equation to zero:

$0 = 2x^2 - 14x + 49 - 169$

$2x^2 - 14x - 120 = 0$

Divide the entire equation by $2$ to simplify:

$x^2 - 7x - 60 = 0$

Step 4: Solving the Quadratic Equation by Factorization

We need to find two numbers that multiply to $-60$ and add to $-7$. These numbers are $-12$ and $5$.

$x^2 - 12x + 5x - 60 = 0$

$x(x - 12) + 5(x - 12) = 0$

$(x - 12)(x + 5) = 0$

Step 5: Determining the Possible Values for x

Setting each factor to zero:

1) $x - 12 = 0 \implies x = 12$

2) $x + 5 = 0 \implies x = -5$

Since the length of a side of a triangle cannot be negative, we discard $x = -5$.

Therefore, the base $x = 12$ cm.

Step 6: Calculating the Altitude

Altitude = $x - 7$

Altitude = $12 - 7 = 5$ cm.

Final Answer: The base of the triangle is 12 cm and the altitude is 5 cm.

Given:

1. The total cost of production on a particular day = ₹ $90$.

2. The cost of production of each article is $3$ more than twice the number of articles produced.

To Find:

1. The number of articles produced on that day.

2. The cost of each article.

Step 1: Defining Variables

Let the number of pottery articles produced on that day be $x$.

According to the problem, the cost of production of each article is $3$ more than twice the number of articles produced.

Therefore, the cost of each article = $(2x + 3)$ rupees.

Step 2: Formulating the Quadratic Equation

The total cost of production is calculated by multiplying the number of articles by the cost per article.

Total Cost = (Number of articles) $\times$ (Cost of each article)

$90 = x(2x + 3)$ [Given that total cost is ₹ 90]

$90 = 2x^2 + 3x$ [Distributive property of multiplication]

Rearranging the terms to form a standard quadratic equation $ax^2 + bx + c = 0$:

$2x^2 + 3x - 90 = 0$

Step 3: Solving the Quadratic Equation by Factorization

To factorize $2x^2 + 3x - 90 = 0$, we need to find two numbers whose product is $a \times c = 2 \times (-90) = -180$ and whose sum is $b = 3$.

The factors of $-180$ that add up to $3$ are $15$ and $-12$.

Splitting the middle term:

$2x^2 + 15x - 12x - 90 = 0$

Grouping the terms:

$(2x^2 + 15x) - (12x + 90) = 0$

$x(2x + 15) - 6(2x + 15) = 0$ [Factoring out common terms]

$(2x + 15)(x - 6) = 0$

Step 4: Finding the Roots

Setting each factor to zero:

1. $2x + 15 = 0 \implies 2x = -15 \implies x = -7.5$

2. $x - 6 = 0 \implies x = 6$

Since the number of articles produced cannot be negative, we discard $x = -7.5$.

Therefore, the number of articles produced is $x = 6$.

Step 5: Calculating the Cost of Each Article

Cost of each article = $2x + 3$

Substitute $x = 6$ into the expression:

Cost = $2(6) + 3$

Cost = $12 + 3 = 15$

The cost of each article is ₹ $15$.

Final Answer: The number of articles produced is 6 and the cost of each article is ₹ 15.