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  1. Asked: In:

    Find the roots of the following equations: x – 1/x = 3, x ≠ 0

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    jony jain

    x - 1/x = 3, x ≠ 0 ⇒ x² - 1 = 3x ⇒ x² - 3x - 1 = 0 For the quadratic equation x² - 3x - 1 = 0, we have a = 1, b = - 3 and c = - 1. Therefore, b² - 4ac = (-3)² - 4 × 1 × (-1) = 9 + 4 = 13 > 0 Hence, x = (3 ± √13)/2 [As x = (-b ±√(b²- 4ac))/2a] Either x = (3 + √13)/2 or x = (3 - √13)/2 Hence, the rRead more

    x – 1/x = 3, x ≠ 0
    ⇒ x² – 1 = 3x
    ⇒ x² – 3x – 1 = 0

    For the quadratic equation x² – 3x – 1 = 0, we have a = 1, b = – 3 and c = – 1.
    Therefore, b² – 4ac = (-3)² – 4 × 1 × (-1) = 9 + 4 = 13 > 0

    Hence, x = (3 ± √13)/2 [As x = (-b ±√(b²- 4ac))/2a]

    Either x = (3 + √13)/2 or x = (3 – √13)/2
    Hence, the roots of the quadratic equation are (3 + √13)/2 and (3 – √13)/2.

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  2. Asked: In:

    The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

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    jony jain

    Let the smaller side = x m Therefore, hypotenuse = x + 60 m So, the longer side = x + 30 m According to the question, (x + 60)² = x² + (x + 30)² ⇒ x² + 120x +3600 = x² + x² + 60x + 900 ⇒ - x² + 60x + 2700 = 0 ⇒ x² - 60 - 2700 = 0 ⇒ x² - 90 + 30x - 2700 = 0 ⇒ x(x - 90) + 30 (x - 90) = 0 ⇒ (x - 90) (xRead more

    Let the smaller side = x m
    Therefore, hypotenuse = x + 60 m
    So, the longer side = x + 30 m
    According to the question, (x + 60)² = x² + (x + 30)²
    ⇒ x² + 120x +3600 = x² + x² + 60x + 900
    ⇒ – x² + 60x + 2700 = 0
    ⇒ x² – 60 – 2700 = 0
    ⇒ x² – 90 + 30x – 2700 = 0
    ⇒ x(x – 90) + 30 (x – 90) = 0
    ⇒ (x – 90) (x + 30) = 0
    ⇒ (x – 90) = 0 or ( x + 30 ) = 0
    Either x = 90 or x = – 30
    But, x ≠ – 30 ,as x is side of field which can’t be negative.
    Therefore, x = 90 and hence the smaller side = 90 m
    So, the longer side = 90 + 30 = 120 m

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  3. Asked: In:

    The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

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    jony jain

    Let the larger number = x Let the smaller number = y Therefore, y² = 8x According to the question, x² - y² = 180 ⇒ x² - 8x = 180 [ As, y² = 8x ] ⇒ x² - 8x - 180 = 0 ⇒ x² - 18x + 10x - 180 = 0 ⇒ x(x - 18) +10 (x -18) = 0 ⇒ (x - 18) (x + 10) = 0 ⇒ (x - 18) = 0 or, (x + 10) = 0 Either x = 18 or x = - 1Read more

    Let the larger number = x
    Let the smaller number = y
    Therefore, y² = 8x
    According to the question, x² – y² = 180
    ⇒ x² – 8x = 180 [ As, y² = 8x ]
    ⇒ x² – 8x – 180 = 0
    ⇒ x² – 18x + 10x – 180 = 0
    ⇒ x(x – 18) +10 (x -18) = 0
    ⇒ (x – 18) (x + 10) = 0
    ⇒ (x – 18) = 0 or, (x + 10) = 0
    Either x = 18 or x = – 10
    But, x ≠ – 10, as x larger of the two numbers So, x = 18
    Therefore, the larger number = 18
    Hence the smaller number = y = √144 = 12

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  4. Asked: In:

    In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

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    jony jain
    • 1
  6. Asked: In:

    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 rupees 90, find the number of articles produced and the cost of each article.

    Best Answer
    jony jain

    Let, the number of article = x Therefore, the cost of one article = 2x + 3 According to question, the total cost = 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 or (x - 6) = 0 Either x = - 15/2 or xRead more

    Let, the number of article = x
    Therefore, the cost of one article = 2x + 3
    According to question, the total cost = 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 or (x – 6) = 0
    Either x = – 15/2 or x = 6
    But, x ≠ – 15/2, as x is a number of articles.
    Therefore, x = 6 and the cost of each articles = 2x + 3 = 2 × 6 + 3 = 15
    Hence, the number of articles = 6 and the cost of each article is Rs 15.

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