Respuesta :
Answer: " x = 1 ± 2i√2 " .
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Step-by-step explanation:
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To solve the given equation:
" [tex]x^{2} -2x = -9[/tex] " ;
for all values of "[tex]x[/tex]" ; by using the "quadratic formula equation" ;
we must first rewrite the given equation in "quadratic format" ;
→ that is; in the format:
→ " [tex]ax^{2} +bx + c= 0[/tex] " ; {[tex]a\neq 0}[/tex]} ;
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So, given:
" [tex]x^{2} -2x = -9[/tex] " ;
→ 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 " ;
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→ which is written in the "quadratic format" ; that is:
→ " [tex]ax^{2} +bx + c= 0[/tex] " ; {[tex]a\neq 0[/tex]} ;
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.}. "
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To solve for all value(s) for "[tex]x[/tex]" in our given equation by using the quadratic equation formula:
→ We use the following quadratic equation formula:
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→ x = [-b ± √(b² − 4ac) ] / [2a] ;
Note that in the "quadratic formula" :
→ " [tex]ax^{2} +bx + c= 0[/tex] " ; "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" .
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So, let us solve for the values for "[tex]x[/tex]" ; by plugging in our known values into the quadratic equation formula:
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→ [tex]x[/tex] = [[tex]-b[/tex] ± [tex]\sqrt{(b^{2}-4ac)} ] / [2a] ;[/tex]
So, to solve for "x" ; plug in "1" for "a"; "-2" for "b" ; "9" for "c" ;
→ So: " [tex](-b) = -(-2) = 2[/tex] " ;
" b² = (-2)² = (-2) * (-2) = 4 " ;
" 4ac = 4 * 1 * 9 = 4 * 9 = 36 ;
" 2a = 2 * 1 = 2 " ;
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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.
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→ Now, we can further simplify:
→ i√32;
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Let us simplify: "√32" ;
→ √32 = √16 * √2 = 4 * √2 ; or, write as 4√2 ;
So: √-32 = i√-32 = i * 4√2 = 4i√2 .
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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:
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→ " x = 1 ± 2i√2 " .
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Hope this answer—and explanation—is helpful!
Wishing you well in your academic endeavors!
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