Anonymous
Anonymous asked in Science & MathematicsMathematics · 2 months ago

Solve each equation by completing the square using the steps in the lesson content.?

x2 + 2x + 5 = 0 ... I got x = -1 and x = -1 - 2i , but don't know if that is right.

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  • 2 months ago
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    Answer: x = -1 + 2i and x = -1 - 2i

    x² + 2x + 5 = 0

    Second term is the key here.

    Remember

    (a + b)² = a² + 2ab + b²

    and recognize that in your equation 2x can be written as 2*x*1 so a=x and b=1

    (x + 1)² = x² + 2x + 1

    What is needed to make this equal to the left side of our equation? Answer: we have to add 4.

     x² + 2x + 5 = (x² + 2x + 1) + 4 =

    = (x + 1)² + 4 = 0

    from here

    (x + 1)² = -4

    square root of both sides

    x + 1 = ±√(-4)

    we'll find this root step by step to make it clear

    x + 1 = ±√(4 * (-1)) = ±2√(-1) = ±2i

    subtract 1 from both sides

    x = -1 ± 2i

    Our two solutions are

    x = -1 + 2i

    x = -1 - 2i

  • 2 months ago

    x² + 2x + 5 => (x + 1)² + 5 - 1

    i.e. (x + 1)² + 4 = 0

    so, (x + 1)² = -4

    Then, x + 1 = ±2i

    Hence, x = -1 ± 2i

    :)>

  • 2 months ago

    x^2 + 2x + 5 = 0

    (x + 1)^2 + 4 = 0

    Complex solutions:

    x = -1 - 2 i

    x = -1 + 2 i

  • 2 months ago

    No.  It's not correct.

    You'd either get 0 complex roots or 2 complex roots.  You would never only get one.

    Let's see what I get:

    x² + 2x + 5 = 0

    Let's start with moving the 5 to the other side:

    x² + 2x = -5

    Now complete the square by adding 1 to both sides (square of half of 2):

    x² + 2x + 1 = -4

    Now we can factor the left side:

    (x + 1)² = -4

    Square root of both sides:

    x + 1 = ±2i

    subtract 1 from both sides:

    x = -1 ± 2i

    One of your roots were correct, but both roots are complex numbers.

    complex and irrational roots in a polynomial are always in conjugate pairs.

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