Respuesta :
The answer is: [A]: "one rational solution" .
_______________________________________________________
Explanation:
_______________________________________________________
Given: 2c² = 16c − 32 ;
Let us rewrite this equation into:
_________________________________________________
"quadratic format" ; or: "ax² + bx + c = 0 ; a ≠ 0 "
__________________________________________________
Given: "2c² = 16c − 32" ; rewrite — substituting "x" for "c" ;
_________________________________________________
→ 2x² = 16x − 32 ;
_______________________________________________________
→ Subtract "16x" ; and add "32" ; to EACH side of the equation;
_______________________________________________________
→ 2x² − 16x + 32 = 16x − 32 − 16x + 32 ; and rewrite:
_______________________________________________________
→ 2x² − 16x + 32 = 0
_______________________________________________________
So, this equation is now in "quadratic format" ; that is:
______________________________________________
→ ax² + bx + c = 0 ; (a ≠ 0) ;
in which: a = 2 ;
b = -16 ;
c = 32 ;
_____________________________________________
So, we have the quadratic equation:
→ 2x² − 16x + 32 = 0 ;
→ Let us divide the ENTIRE EQUATION by "2" ;
_________________________________________________________
→ {2x² − 16x + 32} / 2 = 0 / 2 ;
→ x² − 8x + 16 = 0 ;
→ We can use the quadratic equation formula; but if we can factor the equation; it would be easier. Can we factor it? Yes!
__________________________________________________
→ x² − 8x + 16 = (x − 4) (x − 4) ;
______________________________________________________
→ (x − 4)(x − 4) = 0 ;
_______________________________________________________
Since each "multiplicand" is the same; that is: "(x − 4)" ; AND anything multiplied by "0" is equal to "0" ; we can solve for "x" as follows:
_______________________________________________________
→ (x − 4) = 0 ; Solve for "x" ;
_______________________________________________________
→ x − 4 = 0 ; Add "4" to each side of the equation; to isolate "x" on one side of the equation; and to solve for "x" ;
_______________________________________________________
→ x − 4 + 4 = 0 + 4 ;
→ x = 4 .
{Note: this is the long version; with practice (or inherently), by visual inspection, see that given: (x − 4) = 0 ; x = 4 .}.
_______________________________________________________
So, since we substituted "x" for "c"; we rewrite the answer as:
__________________________________________________
→ c = 4 .
__________________________________________________
Let us check our answer:
__________________________________________________
By plugging in "4" for "c" ; in the original equation:
__________________________________________________
→ 2c² = 16c − 32 ;
__________________________________________________
→ 2*4² =? 16*4 − 32 ?? :
→ 2*16 =? 64 − 32 ??
→ 32 = 32 . Yes!
______________________________________________________
Since there is only one value for "c" ; which is, "4" (positive 4), the answer to the question is: Answer choice: [A]: "one rational solution" .
__________________________________________________________
_______________________________________________________
Explanation:
_______________________________________________________
Given: 2c² = 16c − 32 ;
Let us rewrite this equation into:
_________________________________________________
"quadratic format" ; or: "ax² + bx + c = 0 ; a ≠ 0 "
__________________________________________________
Given: "2c² = 16c − 32" ; rewrite — substituting "x" for "c" ;
_________________________________________________
→ 2x² = 16x − 32 ;
_______________________________________________________
→ Subtract "16x" ; and add "32" ; to EACH side of the equation;
_______________________________________________________
→ 2x² − 16x + 32 = 16x − 32 − 16x + 32 ; and rewrite:
_______________________________________________________
→ 2x² − 16x + 32 = 0
_______________________________________________________
So, this equation is now in "quadratic format" ; that is:
______________________________________________
→ ax² + bx + c = 0 ; (a ≠ 0) ;
in which: a = 2 ;
b = -16 ;
c = 32 ;
_____________________________________________
So, we have the quadratic equation:
→ 2x² − 16x + 32 = 0 ;
→ Let us divide the ENTIRE EQUATION by "2" ;
_________________________________________________________
→ {2x² − 16x + 32} / 2 = 0 / 2 ;
→ x² − 8x + 16 = 0 ;
→ We can use the quadratic equation formula; but if we can factor the equation; it would be easier. Can we factor it? Yes!
__________________________________________________
→ x² − 8x + 16 = (x − 4) (x − 4) ;
______________________________________________________
→ (x − 4)(x − 4) = 0 ;
_______________________________________________________
Since each "multiplicand" is the same; that is: "(x − 4)" ; AND anything multiplied by "0" is equal to "0" ; we can solve for "x" as follows:
_______________________________________________________
→ (x − 4) = 0 ; Solve for "x" ;
_______________________________________________________
→ x − 4 = 0 ; Add "4" to each side of the equation; to isolate "x" on one side of the equation; and to solve for "x" ;
_______________________________________________________
→ x − 4 + 4 = 0 + 4 ;
→ x = 4 .
{Note: this is the long version; with practice (or inherently), by visual inspection, see that given: (x − 4) = 0 ; x = 4 .}.
_______________________________________________________
So, since we substituted "x" for "c"; we rewrite the answer as:
__________________________________________________
→ c = 4 .
__________________________________________________
Let us check our answer:
__________________________________________________
By plugging in "4" for "c" ; in the original equation:
__________________________________________________
→ 2c² = 16c − 32 ;
__________________________________________________
→ 2*4² =? 16*4 − 32 ?? :
→ 2*16 =? 64 − 32 ??
→ 32 = 32 . Yes!
______________________________________________________
Since there is only one value for "c" ; which is, "4" (positive 4), the answer to the question is: Answer choice: [A]: "one rational solution" .
__________________________________________________________
