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Natali5045456 [20]
4 years ago
6

Use the quadratic formula to solve equation x²-2x=-9

Mathematics
1 answer:
andreev551 [17]4 years ago
8 0

Answer:  "  x  =  1  ±  2i√2 " .

_________________________________

Step-by-step explanation:

_________________________________

To solve the given equation:

  " x^{2} -2x = -9 " ;

for all values of "x" ; by using the "quadratic formula equation" ;

we must first rewrite the given equation in "<u>quadratic format</u>" ;

    →  that is;  in the format:

    →  " ax^{2} +bx + c= 0 " ;  {a\neq 0}} ;

________________________________

So, given:

 " x^{2} -2x = -9 " ;

→  We must add "9" to Each Side of the equation; to get the equation into "quadratic format" ;  as follows:

          →  x² − 2x + 9 = -9 + 9 ;

to get:

          →  " x² − 2x + 9 = 0 "  ;

___________________________________

          →  which is written in the "quadratic format" ;  that is:

    →  " ax^{2} +bx + c= 0 " ; {a\neq 0} ;

in which:

"a = 1" ;  [the "implied coefficient;  "1" ; since "1" multiplied by "any value" ;

                                                        results in "that same value" ;

                                  → {Note this this is known as the "identity property"

                                                                                of multiplication.}.    

  b = -2 ;

  c =  9 ;  

and since:  " a = 1 " ;  We know that: " {a≠0.}. "

_______________________________________

To solve for all value(s) for "x" in our given equation by using the quadratic equation formula:

   →   We use the following quadratic equation formula:    

________________________________________

   →  x = [-b ± √(b² − 4ac) ] / [2a] ;

Note that in the "quadratic formula" :

   →  " ax^{2} +bx + c= 0 " ;   "a ≠ 0" ;  

   →  Since we divide by "2a" {refer to the "denominator"];

         → and if:  "a" were to equal "0" ; then the "denominator" would equal:  "2a = 2*a = 2*0 = 0" ; and we cannot divide by "0" .

______________________________________________

So, let us solve for the values for "x" ; by plugging in our known values into the quadratic equation formula:

______________________________________________

   →  x = [-b ± \sqrt{(b^{2}-4ac)} ] / [2a] ;

So, to solve for "x" ; plug in "1" for "a";  "-2" for "b" ;  "9" for "c" ;

  →  So:  " (-b) = -(-2) = 2  " ;  

            "  b²  = (-2)² = (-2) * (-2) = 4 " ;

            " 4ac = 4 * 1 * 9 = 4 * 9 = 36 ;

            " 2a = 2 * 1 = 2 " ;

__________________________________________

So:  

→  x =  (2 ± √(4−36) / 2 ;

Note:  4−36 = -32 ;

So:

→  x = (2 ± √-32) / 2 ;

Note: √-32 ;  can be written as " i * √-32" ;  or simply:  "i√-32" ; since "i" is an 'imaginary number" that can represent the imaginary number:  "√-1" ;  

 →   since:  "√-32 = √-1 * √32" ;  if:  "√-1" really existed.

__________________________________________

→ Now, we can further simplify:

→  i√32;

_________________________________________

Let us simplify:  "√32" ;

→  √32 = √16 * √2 = 4 * √2 ;  or, write as 4√2 ;  

 So:  √-32 = i√-32 = i * 4√2 = 4i√2 .

_________________________________________

We have, from above:  

→  x = (2 ± √-32) / 2 ;

Rewrite—by substituting "4i√2" ;  in lieu of:  "√-32" ; as following

→  x = (2 ± 4i√2) / 2 ;

We are dividing the numerator by "2" ; so we can further simplify:

to get:

________________________________________

→   "  x = 1  ±  2i√2 " .

________________________________________

Hope this answer—and explanation—is helpful!

  Wishing you well in your academic endeavors!

________________________________________

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