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  1. Asked: In:

    Find two consecutive positive integers, sum of whose squares is 365.

    Best Answer
    jony jain

    Let the first number = x Therefore, second number = x + 1 according to the questions, x² + (x + 1)² = 365 ⇒ x² + x² + 2x + 1 = 365 ⇒ 2x² + 2x - 364 = 0 ⇒ x² + x - 182 = 0 ⇒ x² - 13x + 14x + 182 = 0 ⇒ x(x - 13) + 14 (x - 13) = 0 ⇒ (x - 13) (x + 14) = 0 ⇒ (x - 13) = 0 or (x + 14) = 0 ⇒ Either x = 13 oRead more

    Let the first number = x Therefore, second number = x + 1
    according to the questions, x² + (x + 1)² = 365
    ⇒ x² + x² + 2x + 1 = 365 ⇒ 2x² + 2x – 364 = 0
    ⇒ x² + x – 182 = 0 ⇒ x² – 13x + 14x + 182 = 0
    ⇒ x(x – 13) + 14 (x – 13) = 0 ⇒ (x – 13) (x + 14) = 0
    ⇒ (x – 13) = 0 or (x + 14) = 0 ⇒ Either x = 13 or x = – 14
    Hence, the two consecutive positive integers are 13 and 14.

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  2. Asked: In:

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

    Best Answer
    jony jain

    Let the base = x cm Therefore the height = x - 7 cm Given that : hypotenuse = 13 cm Using Pythagoras theorem, x² + (x - 7)² = 13² ⇒ x² + x² - 14x + 49 = 169 ⇒ 2x² - 14x - 120 = 0 ⇒ x² - 7x - 60 = 0 ⇒ x² - 12x + 5x - 60 = 0 ⇒ x(x - 12) + 5 (x - 12) = 0 ⇒ (x - 12) (x + 5) = 0 ⇒ (x -12) = 0 or (x + 5)Read more

    Let the base = x cm Therefore the height = x – 7 cm
    Given that : hypotenuse = 13 cm
    Using Pythagoras theorem, x² + (x – 7)² = 13²
    ⇒ x² + x² – 14x + 49 = 169 ⇒ 2x² – 14x – 120 = 0
    ⇒ x² – 7x – 60 = 0 ⇒ x² – 12x + 5x – 60 = 0
    ⇒ x(x – 12) + 5 (x – 12) = 0 ⇒ (x – 12) (x + 5) = 0
    ⇒ (x -12) = 0 or (x + 5) = 0
    Either x = 12 or x = – 5
    But x ≠ – 5, as x is side of triangle.
    Therefore, x = 12 and the second side = x – 7 = 12 – 7 = 5
    Hence, the other two sides are 5 cm and 12 cm.

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  3. Asked: In:

    Find two numbers whose sum is 27 and product is 182.

    Best Answer
    jony jain

    Let the first number = x Therefore, the second number = 27 - x According to the question, Product, = 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 or (x - 14) = 0 Either x = 13 or x = 14 Hence, theRead more

    Let the first number = x
    Therefore, the second number = 27 – x
    According to the question,
    Product, = 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 or (x – 14) = 0
    Either x = 13 or x = 14
    Hence, the two required numbers are 13 and 14

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  4. Asked: In:

    Find the roots of the following quadratic equations by factorization: x² – 3x – 10 = 0

    Best Answer
    jony jain

    x² - 3x - 10 = 0 solving the quadratic equations, we got x² - 3x - 10 = 0 ⇒ x² - 5x + 2x + 10 = 0 ⇒ x(x -5) + 2x (x - 5) = 0 ⇒ (x - 5) (x + 2) = 0 ⇒ (x - 5) = 0 or (x + 2) = 0 Either x = 5 or x = -2 Hence, the roots the given quadratic equations are 5 and -2

    x² – 3x – 10 = 0
    solving the quadratic equations, we got
    x² – 3x – 10 = 0
    ⇒ x² – 5x + 2x + 10 = 0
    ⇒ x(x -5) + 2x (x – 5) = 0
    ⇒ (x – 5) (x + 2) = 0
    ⇒ (x – 5) = 0 or (x + 2) = 0
    Either x = 5 or x = -2
    Hence, the roots the given quadratic equations are 5 and -2

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