Respuesta :
Answer: [B]: Two (2) real solutions .
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Explanation:
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Given: -7x² − 11x − 2 = 0 ;
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We see that it is in "quadratic format" ; that is:
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" ax² + bx + c = 0 ; a ≠ 0 ; "
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Multiply the entire equation by "-1" ; to get rid of the "negative number" :
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→ -1 * {-7x² − 11x − 2 = 0} ;
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to get: → 7x² + 11x + 2 = 0 ;
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This expression is written in the "quadratic format";
ax² + bx + c = 0 ; in which: a = 7 ; b = 11; c = 2 ;
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The expression cannot be "factored"; so, we can solve for "x" ; by using the "quadratic equation formula" ;
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x = {-b ± √(b² − 4ac)} / {2a} ;
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Let us solve for: "(2a)" : 2a = 2 * a = 2 * 7 = 14 ;
Let us solve for: "(b² − 4ac)": 11² − 4*7*2 = 121 − 56 = 65;
→ √(b² − 4ac) = √65
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→ " -b" = -11 ;
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So, x = (-11 ± √65) / 14 ;
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There are TWO (2) solutions:
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Solution 1) x = (-11 + √65) / 14 = - 0.2098387322643893071; AND:
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Solution 2) x = (-11 − √65) / 14 = -1.3615898391641821214 ;
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So, there are TWO (2) real solutions.
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__________________________________________________________
Explanation:
__________________________________________________________
Given: -7x² − 11x − 2 = 0 ;
__________________________________________________________
We see that it is in "quadratic format" ; that is:
__________________________________________________________
" ax² + bx + c = 0 ; a ≠ 0 ; "
__________________________________________________________
Multiply the entire equation by "-1" ; to get rid of the "negative number" :
__________________________________________________________
→ -1 * {-7x² − 11x − 2 = 0} ;
__________________________________________________________
to get: → 7x² + 11x + 2 = 0 ;
__________________________________________________________
This expression is written in the "quadratic format";
ax² + bx + c = 0 ; in which: a = 7 ; b = 11; c = 2 ;
__________________________________________________________
The expression cannot be "factored"; so, we can solve for "x" ; by using the "quadratic equation formula" ;
__________________________________________________________
x = {-b ± √(b² − 4ac)} / {2a} ;
______________________________________
Let us solve for: "(2a)" : 2a = 2 * a = 2 * 7 = 14 ;
Let us solve for: "(b² − 4ac)": 11² − 4*7*2 = 121 − 56 = 65;
→ √(b² − 4ac) = √65
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→ " -b" = -11 ;
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So, x = (-11 ± √65) / 14 ;
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There are TWO (2) solutions:
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Solution 1) x = (-11 + √65) / 14 = - 0.2098387322643893071; AND:
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Solution 2) x = (-11 − √65) / 14 = -1.3615898391641821214 ;
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So, there are TWO (2) real solutions.
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