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LDP
Lv 6
LDP asked in Science & MathematicsMathematics · 10 years ago

Maths expressions and equations (Secondary school)?

I have just finished a 70 page Maths booklet (that was fun...) but there are some questions that I just could not answer. All help on ANY of the questions would be great and I would prefer working out so I can understand.

http://i724.photobucket.com/albums/ww244/dark_phan...

1 Answer

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  • POMPOM
    Lv 5
    10 years ago
    Favorite Answer

    Q1

    8x - x² = 16 + 8x - x² - 16 = 16 - (x² - 8x + 16) = 16 - (x² - 2.x.4 + 4²) = 16 - (x - 4)²

    So, p = 16, q = 4

    The above expression is maximum when the (x - 4)² part vanishes.

    So, the maximum value of the given expression occurs when x = 4 & the maximum value is 16.

    Q2.

    x² + y² = 25 ........... (1)

    y = 2x - 2 ...............(2)

    Put y = 2x - 2 in (1)

    x² + (2x - 2)² = 25

    ⇒ x² + 4x² - 8x + 4 = 25

    ⇒ 5x² - 8x - 21 = 0

    ⇒ 5x² - 15x + 7x - 21 = 0

    ⇒ 5x(x - 3) + 7(x - 3) = 0

    ⇒ (x - 3)(5x + 7) = 0

    ⇒ x = 3, -7/5

    When x = 3, y = 2x - 2 = 4

    When x = -7/5, y = 2x - 2 = -24/5

    So, solutions are (x = 3, y = 4) & (x = -7/5, y = -24/5)

    Q3.

    7/(x+2) + 1/(x-1) = 4

    ⇒ [7(x-1) + (x+2)]/[(x+2)(x-1)] = 4

    ⇒ [7x - 7 + x + 2)]/[(x+2)(x-1)] = 4

    ⇒ 8x - 5 = 4(x² + x -2)

    ⇒ 4(x² + x -2) - 8x + 5 = 0

    ⇒ 4x² + 4x - 8 - 8x + 5 = 0

    ⇒ 4x² - 4x - 3 = 0

    ⇒ 4x² - 6x + 2x - 3 = 0

    ⇒ 2x(2x - 3) + (2x - 3) = 0

    ⇒ (2x - 3)(2x + 1) = 0

    ⇒ x = 3/2, -1/2

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