Kaliyuga Ekalavya

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (iv) Question: Find the roots of the following quadratic equations by factorisation : 2x 2 – x + (1/8) = 0 Given: The quadratic equation: 2x 2 – x + 1 8 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 – x + 1 8 = 0. First, clear the fra...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 6 Question: 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. Given: Cost of production of each article = 3 + 2 × (Number of articles produced). Total cost of production = Rs 90. To Find: The number of articles produced and the cost of each article. Formula: WKT, Total Cost of Production = Number of Articles Produced × Cost of Production of Each Article. WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                          ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (v) Question: Find the roots of the following quadratic equations by factorisation : 100x 2 – 20x + 1 = 0 Given: The quadratic equation: 100x 2 – 20x + 1 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 100x 2 – 20x + 1 = 0. Here, a = 100, b = -20, c = 1. Product (a × c)...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (iii) Question: Find the roots of the following quadratic equations by factorisation: √2x 2 + 7x + 5√2 = 0 Given: The quadratic equation: √2x 2 + 7x + 5√2 = 0 To Find: The values of x (roots) for the given quadratic equation. Formula: WKT, For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given quadratic equation is √2x 2 + 7x + 5√2 = 0. Comparing the given quadr...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (ii) Question: Find the roots of the quadratic equation:Find the roots of the quadratic equation 2x 2 + x – 6 = 0 . Given: The quadratic equation: 2x 2 + x – 6 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 + x – 6 = 0. Here, a = 2, b = 1, c = -6. Product (a × c) ...

NCERT Class X Chapter 4: Quadratic Equation Exercise 4.2 Question 1 (i) Question: Find the roots of the following quadratic equations by factorisation: x 2 – 3x – 10 = 0 Given: The quadratic equation: x 2 – 3x – 10 = 0 To Find: The roots of the equation by factorization. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is x 2 – 3x – 10 = 0. Here, a = 1, b = -3, c = -10. Product (a × c) = 1 × (-1...

NCERT Class X Chapter 4: Quadratic Equation Example 6 Question: Charity trust decides to build a prayer hall having a carpet area of 300 square metres with its length one metre more than twice its breadth. What should be the length and breadth of the hall?. Given: Carpet area of the prayer hall = 300 m 2 Length = 1 metre more than twice its breadth. To Find: The length of the hall. The breadth of the hall Formula: Area of a rectangle = Length × Breadth. For a quadratic equation ax 2 + bx + c = 0, the roots are given by the quadratic formula: x = -b ± √(Δ) 2a Δ = b 2 - 4ac Solution: Let the breadth of the hall (in metres) = x According to the given condition, the length is one metre more than twice its breadth. ⇒ Length (in metres) = ( 1 + 2x ) = ( 2x + 1 ) Area of the hall = Length × Breadth. ⇒ 300 = (2x + 1) × x ⇒ 300 = 2x 2 + x Group the terms on one side of t...

NCERT Class X Chapter 4: Quadratic Equation Example 5 Question: Find the roots of the quadratic equation:Find the roots of the quadratic equation 3x 2 - 2√6x - 2 = 0 . Given: The quadratic equation : 3x 2 - 2√6x - 2 = 0 To Find: The roots of the equation. Formula: WKT, 2√ax = √ax + √ax Solution: The given equation is 3x 2 - 2√6x - 2 = 0. Split the middle term: using the formula 2√ax = √ax+ √ax ⇒ 3x 2 - 2√6x - 2 = 3x 2 - (√6x + √6x) - 2 ⇒ 3x 2 - 2√6x - 2 = 3x 2 - √6x - √6x - 2 Grouping common factors in 3x 2 - √6x - √6x - 2 we have ⇒ (3x 2 - √6x) + (- √6x - 2) = √3x(√3x-√2) -√2(√3x-√2) ⇒ 3x 2 - 2√6x - 2 = (√3x-√2) (√3x-√2) Since 3x 2 - 2√6x - 2 = 0, (√3x-√2) (√3x-√2) = 0 Case 1: ⇒ (√3x-√2) = 0 ⇒ √3x = √2 ⇒ x = √2 √3 Case 2: ⇒ (√3x-√2) = 0 ⇒ √3x = √2 ⇒ x = √2 √3 Therefor the two roots are: x = √2 √3...

NCERT Class X Chapter 4: Quadratic Equation Example 4 Question: Find the roots of the quadratic equation 6x 2 – x – 2 = 0. Given: The quadratic equation: 6x 2 – x – 2 = 0 To Find: The roots of the equation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 6x 2 – x – 2 = 0. Here, a = 6, b = -1, c = -2. Product (a × c) = 6 × (-2) = -12. Sum (b) = -1. We need to find two numbers whose ...

NCERT Class X Chapter 4: Quadratic Equation Example 3 Question: Find the roots of the equation 2x 2 – 5x + 3 = 0, by factorisation. Given: The quadratic equation: 2x 2 – 5x + 3 = 0 To Find: The roots of the equation by factorisation. Formula: For a quadratic equation ax² + bx + c = 0 , we can find two numbers m and n such that:                               ax² + bx + c = ax² + mx + nx + c where m × n = a × c m + n = b Cases: If (a × c > 0) and b < 0 , both m and n are negative . If (a × c > 0) and b > 0 , both m and n are positive . If (a × c < 0) , the numbers m and n have opposite signs . Solution: The given equation is 2x 2 – 5x + 3 = 0. Here, a = 2, b = -5, c = 3. Product (a × c) = 2 × 3 = 6. Sum (b) = -5. We need to...