What number should be added to both sides of the equation to complete the square? x2 – 6x = 5

What number should be added to both sides of the equation to complete the square?

x2 – 6x = 5

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This Post Has 8 Comments

  1. 9

    Step-by-step explanation:

    Given

    x² - 6x = 5

    To complete the square

    add ( half the coefficient of the x- term )² to both sides

    x² + 2(- 3)x + 9 = 5 + 9, that is

    (x - 3)² = 14

  2. 9 should be added to complete the square

    Step-by-step explanation:

    1. Write the equation in a way that the constants are in the right side while the terms with x are on the left. 

    x^2 – 6x = 5

    2. Make sure that the coefficient of the x^2 term is 1.

    x^2 – 6x = 5

    3. Adding a term to both sides that will complete the square in the left side. This is done by dividing the coefficient of the x term by 2 and squaring it. Note: The same amount should be added to the right side to balance the equation.

    x^2 – 6x + 9 = 5 + 9

    (x-3)^2 = 14

    Therefore, 9 should be added to both sides to complete the square.

  3. I think 9 should be added to both sides.

    cofficient of x = 6

    half of it = 6/2 = 3

    square the 3

    to give 3 squared = 3*3 9

  4. 9

    Step-by-step explanation:

    To make it a Perfect Square Trinomial, you can square root 9 and multiply it to get 6, therefore 9 is correct

  5. First of all, it is better to note that x2 is not equal to x² .In mathematics x2 is equal to 2nd x or 2x i.e. 2 times x=2*x=x*2.

    I don't know what this question means ! But if it means to just confine x2 and -6x in a square or to make them part of a square then it could be done as follows.

    Assume x2 as x² as per your needs as case may be.

    Now use the identity

    (a² - 2a×b + b²) =(a - b )²

    Here , in LHS

    a appears twice as in a² and in 2a×b.

    So for given expression (x2 – 6x = 5) or (x² – 6x = 5) or (x² - 6x -5 =0)

    Assume x=a

    Then,

    6x =2*x*3 =2*a*b

    => 3 = b

    Thus

    (x² - 6x -5 =0)

    => x² - 2*x*3 + 3² -3² -5 =0

    => ( x² - 2*x*3 + 3² ) - 3² -5 =0

    => (x - 3)² - 3² = 5

    => (x - 3)² = 5 + 3²

    => (x - 3)² = 5 + 9 = 14 = ( ±√(14) )²

    Now comparing

    (x² – 6x = 5) and (x - 3)² = 5 + 9,

    it can be said that number 9 should be added to both sides of (x² – 6x = 5) or (x² – 6x) = ( 5 ) to make ( x² ) and (-6x ) as the constituting parts of the square (x-3)² .

    I hope it helps.

    Thank you  for patiently reading it.

  6. 9

    Step-by-step explanation:

    x^2 - 6x = 5

    Take the coefficient of x

    -6

    Divide it by 2

    -6/2 = -3

    Square it

    (-3)^2 =9

    Add 9 to each side

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