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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}

Sagot :

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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

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.

[tex] \small \sf \sqrt{(x + 5) {}^{2} } = \sqrt{49} [/tex]

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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