How do I complete the squares on these problems?

Please help me with these three problems:

x^2-3x-28=0

x^2+6x=-41

2p^2=6p-20

How do I do these?

I need to complete the squares. Please help!

Can someone show me the work somehow?

4 Answers

Relevance
  • n00bz
    Lv 5
    9 years ago
    Favorite Answer

    I can give you the first ...

    x^2-3x-28=0

    x^2-7x+4x-28=0

    x(x-7)+4(x-7)=0

    (x+4)(x-7)=0

    x=-4

    x=7

  • 9 years ago

    Problem 1:

    x^2 - 3x - 28 = 0

    Rewrite it as:

    x^2 - 3x = 28

    Add square of half of the coefficient of the middle term to both sides of equation, get:

    x^2 - 3x + (-3/2)^2 = 28 + (-3/2)^2

    When simplified the LHS and the RHS are:

    (x - 3/2)^2 = 121/4

    Therefore, the LHS is completed as a square.

    Problem 2:

    x^2 + 6x = -41

    Add square of half of the coefficient of the middle term to both sides of equation, get:

    x^2 + 6x + (6/2)^2 = -41 + (6/2)^2

    i.e

    x^2 + 6x + 9 = -41 + 9

    Simplify both LHS and RHS

    (x + 3)^2 = -32

    Also, the LHS is completed as a square.

    Problem 3:

    2p^2 = 6p - 20

    Rewrite this as:

    2p^2 - 6p = -20

    Divide all by 2, get:

    p^2 - 3p = -10

    Need to add the square of half of the coefficient of the middle term to both sides of equation, get:

    p^2 - 3p + (-3/2)^2 = -10 + (-3/2)^2

    By simplifying LHS and RHS differently, get:

    (p - 3/2)^2 = -31/4

    Check, the LHS is completed as a square.

  • 9 years ago

    x^2-3x-28=0

    x^2-7x+4x-28=0

    x(x-7)+4(x-7)=0

    (x+4)(x-7)=0

    either

    x+4 =0 i.e x =-4

    or x-7 =0 i.e x=7

    answer x = -4 or 7

  • Chris
    Lv 7
    9 years ago

    x^2 -3x - 28 = 0

    First, push the constant to the right side by adding 28 to each side. That gets it out of the way.

    x^2 - 3x = 28

    Now to complete the square, take half the x coefficient, square it, and add the result to both sides.

    x^2 - 3x + (-3/2)^2 = 28 + (-3/2)^2

    x^2 - 3x + 9/4 = 28 + 9/4

    x^2 - 3x + 9/4 = 121/4

    (x - 3/2)^2 = 121/4

    Now take the square root of both sides.

    x - 3/2 = +/- sqrt(121/4)

    x - 3/2 = +/- 11/2

    x = 3/2 +/- 11/2

    x = -8/2 or x = 14/2

    x = -4 or x = 7

    x^2 + 6x = -41

    x^2 + 6x + 3^2 = -41 + 3^2

    x^2 + 6x + 9 = -41 + 9

    x^2 + 6x + 9 = -32

    (x + 3)^2 = -32

    x + 3 = +/- sqrt(-32)

    x + 3 = +/- sqrt(-1 * 16 * 2)

    x + 3 = +/- 4i sqrt(2)

    x = -3 +/- 4i sqrt(2)

    x = -3 - 4i sqrt(2) or x = -3 - 4i sqrt(2)

    2p^2 = 6p - 20

    p^2 = 3p - 10

    p^2 - 3p = -10

    p^2 - 3p + (-3/2)^2 = -10 + (-3/2)^2

    p^2 - 3p + 9/4 = -10 + 9/4

    (p - 3/2)^2 = -31/4

    p - 3/2 = +/- sqrt(-31/4)

    p = 3/2 +/- sqrt(-1 * 31) / 2

    p = 3/2 +/- i sqrt(31) / 2

    p = 3/2 + i sqrt(31) or p = 3/2 - i sqrt(31)

Still have questions? Get your answers by asking now.