Solve x2 + 10x = 24 by completing the square. Which is the solution set of the equation? (negative 5 minus StartRoot 34 EndRoot

Question

Solve x2 + 10x = 24 by completing the square. Which is the solution set of the equation?

(negative 5 minus StartRoot 34 EndRoot comma negative 5 + Startroot 34 EndRoot)
(negative 5 minus StartRoot 29 EndRoot comma negative 5 + StartRoot 29 EndRoot)
{–12, 2}
{–2, 12}

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Eirian 4 years 2021-08-05T17:05:56+00:00 2 Answers 41 views 0

Answers ( )

  1. thienthi
    0
    2021-08-05T17:06:58+00:00
    Reply

    Answer:

    (-12,2)

    Step-by-step explanation:

    x^2 + 10x = 24

    x^2 + 10x + (10/2)^2 = 24 + (10/2)^2

    10/2 = 5

    5^2 = 25

    x^2 + 10x + 25 = 24 + 25

    x^2 + 10x + 25 = 49

    (x + 5)^2 = 49                 Take the square root of both sides

    (x + 5) = sqrt(49)

    x + 5 = +/- 7

    x = +/- 7 – 5

    x = +7 – 5 = 2

    x = -7 – 5 = -12

  2. Delwyn
    0
    2021-08-05T17:07:24+00:00
    Reply

    Answer:

    { -12 , 2}

    Step-by-step explanation:

    x² + 10x = 24

    In order to complete the square, the equation must first be in the form x² + bx =c.

    • x² + 10x = 24

    Divide 10, the coefficient of the x term, by 2 to get 5. Then add the square of 5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

    • x² + 10x + 5² = 24 + 5²

    expand exponents.

    • x² + 10x + 25 = 24 + 25

    Add 24 and 25

    • x² + 10x + 25 = 49

    Factor x² + 10x + 25. In general, when x² + bx + c is a perfect square, it can always be factored as ( x + b/2)².

    • ( x + 5 )² =49

    Take the square root of both sides of the equation.

    \small \sf \sqrt{(x + 5) {}^{2} } = \sqrt{49}

    simplify

    • x + 5 = 7
    • x + 5 = +/- 7

    Subtract 5 from both sides.

    x + 5 – 5 = 7 – 5

    • x = 2

    x + 5 – 5 = +/- 7 -5

    • x = -7 – 5 = -12

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )