[tex]Solve 2x^{2}+20x+8=0 completing the squarex=-5+\sqrt{21} x=-5-\sqrt{21}x=5+\sqrt{17}. x=5-\sqrt{17}x=5+\sqrt{21} x=5-\sqrt{21}x=-5+\sqrt{17} x=-5-\sqrt{17}[/tex]

Respuesta :

Answer:

  (a)  x = -5 ± √21

Step-by-step explanation:

You want the solution to 2x² +20x +8 = 0 using the "completing the square" method.

Completing the square

It can be less confusing to divide the equation by the leading coefficient and separate the variable terms from the constant term:

  2x² +20x +8 = 0

  x² +10x +4 = 0 . . . . . . . divide by 2

  x² +10x = -4 . . . . . . . . subtract 4

To complete the square, we add the square of half the x-coefficient:

  x² +10x +25 = -4 +25 . . . . . . . (10/2)² = 5² = 25

  (x +5)² = 21

To find the solution, we take the square root of both sides:

  x +5 = ±√21

  x = -5 ±√21 . . . . . . subtract 5

The solution is x = -5+√21 and x = -5-√21, matching the first choice.