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bhola parsad

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  1. Asked: December 30, 2020In: Class 10 Maths

    Find the roots of the following quadratic equations, if they exist, by the method of completing the square: 2 x² – 7x + 3 = 0

    bhola parsad
    Added an answer on January 10, 2023 at 7:31 am

    2x2 – 7x + 3 = 0 Dividing both side by 2 x² - 7/2 x + 3/2 = 0 ⇒ x² - 7/2 x = - 3/2 Adding [1/2(7/2)]² on both the sides, we get x² - 7/2 x + (7/4)² = - 3/2 + (7/4)² [(as x = -b±√(b²- 4ac)]/2a] ⇒ (x - 7/4)² = - 3/2 + 49/16 ⇒ (x - 7/4)² = - 24 + 49/16 ⇒ (x - 7/4)² = 25/16 ⇒ x - 7/4 = ± 5/4 Either x -Read more

    2×2 – 7x + 3 = 0
    Dividing both side by 2
    x² – 7/2 x + 3/2 = 0 ⇒ x² – 7/2 x = – 3/2
    Adding [1/2(7/2)]² on both the sides, we get
    x² – 7/2 x + (7/4)² = – 3/2 + (7/4)² [(as x = -b±√(b²- 4ac)]/2a]
    ⇒ (x – 7/4)² = – 3/2 + 49/16 ⇒ (x – 7/4)² = – 24 + 49/16
    ⇒ (x – 7/4)² = 25/16
    ⇒ x – 7/4 = ± 5/4
    Either x – 7/4 = 5/4 or x – 7/4 = – 5/4
    ⇒ x = 5/4 + 7/4 or x – 5/4 + 7/4
    ⇒ x = 5+7/4 = 12/4 = 3 x 5+7/4 = 2/4 = 1/2
    Hence, the roots of the quadratic equation are 3 and 1/2.

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  2. Asked: January 4, 2023In: Class 10 Maths

    Check whether the following is quadratic equations : x² – 2x = (–2) (3 – x)

    bhola parsad
    Added an answer on January 10, 2023 at 7:10 am

    x2 – 2x = (–2) (3 – x) simplifying the given equation, we get x2 – 2x = (–2) (3 – x) ⇒ x² - 2x = - 6 + 2x ⇒ x² - 4x + 6 = 0 or x² - 4x + 6 = 0 This is an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.

    x2 – 2x = (–2) (3 – x)
    simplifying the given equation, we get
    x2 – 2x = (–2) (3 – x)
    ⇒ x² – 2x = – 6 + 2x
    ⇒ x² – 4x + 6 = 0
    or x² – 4x + 6 = 0
    This is an equation of type ax² + bx + c = 0.
    Hence, the given equation is a quadratic equation.

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  3. Asked: January 4, 2023In: Class 10 Maths

    Check whether the following is quadratic equation: (x – 2)(x + 1) = (x – 1)(x + 3)

    bhola parsad
    Added an answer on January 10, 2023 at 5:42 am

    (x – 2)(x + 1) = (x – 1)(x + 3) Simplifying the given equation, we get (x – 2)(x + 1) = (x – 1)(x + 3) ⇒ x² - 2x + x - 2 = x² - x +3x - 3 ⇒ - 3x + 1 = 0 or 3x - 1 = 0 This is not an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.

    (x – 2)(x + 1) = (x – 1)(x + 3)
    Simplifying the given equation, we get
    (x – 2)(x + 1) = (x – 1)(x + 3)
    ⇒ x² – 2x + x – 2 = x² – x +3x – 3
    ⇒ – 3x + 1 = 0
    or 3x – 1 = 0
    This is not an equation of type ax² + bx + c = 0.
    Hence, the given equation is a quadratic equation.

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  4. Asked: January 4, 2023In: Class 10 Maths

    Check whether the following is quadratic equation: x² + 3x + 1 = (x – 2)²

    bhola parsad
    Added an answer on January 10, 2023 at 4:32 am

    x² + 3x + 1 = (x – 2)² simplifying the given equation, we get x² + 3x + 1 = (x – 2)² ⇒ x² + 3x + 1 = x² - 4x + 4 ⇒ 7x - 3 = 0 or 7x - 3 = 0 This is not an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.

    x² + 3x + 1 = (x – 2)²
    simplifying the given equation, we get
    x² + 3x + 1 = (x – 2)²
    ⇒ x² + 3x + 1 = x² – 4x + 4
    ⇒ 7x – 3 = 0
    or 7x – 3 = 0
    This is not an equation of type ax² + bx + c = 0.
    Hence, the given equation is a quadratic equation.

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  5. Asked: January 4, 2023In: Class 10 Maths

    Check whether the following is quadratic equation: (2x – 1)(x – 3) = (x + 5)(x – 1)

    bhola parsad
    Added an answer on January 10, 2023 at 4:32 am

    (2x – 1)(x – 3) = (x + 5)(x – 1) Simplifying the given equation, we get (2x – 1)(x – 3) = (x + 5)(x – 1) ⇒ 2x² - x - 6x + 3 = x² + 5x - x - 5 ⇒ x² - 11x + 8 = 0 or x² - 11x + 8 = 0 This is an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.

    (2x – 1)(x – 3) = (x + 5)(x – 1)
    Simplifying the given equation, we get
    (2x – 1)(x – 3) = (x + 5)(x – 1)
    ⇒ 2x² – x – 6x + 3 = x² + 5x – x – 5
    ⇒ x² – 11x + 8 = 0
    or x² – 11x + 8 = 0
    This is an equation of type ax² + bx + c = 0.
    Hence, the given equation is a quadratic equation.

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