Answer: [B]: There are "2" (two) real solutions.
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Explanation:
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Given: -11x² − 10x = -1 ;
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Let us see if we can write this in "quadratic format" ; that is:
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ax² + bx + c = 0 ; a ≠ 0 ;
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So, given: -11x² − 10x = -1 ;
Add "1" to EACH side of the equation;
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-11x² − 10x + 1 = -1 + 1 ;
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to get: -11x² − 10x + 1 = 0 ;
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Now, let us multiply the ENTIRE equation by "-1";
to get rid of the "negative sign" ;
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-1 * { -11x² − 10x + 1 = 0 } ;
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11x² + 10x − 1 = 0 ;
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This equation is written in quadratic format:
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" ax² + bx + c = 0 " ; in which: a = 11 ; b = 10 ; c = -1 ;
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Can the equation: " 11x² + 10x − 1 = 0 " ; be factored? Yes! ;
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→ (x + 1) (11x − 1) = 0 ; There are two solutions;
the equation holds true when EITHER of the TWO MULTIPLICANDS ; or both of them, are equal to "0"; since anything multiplied by "0" equals "0" .
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So, (x + 1) = 0 ;
Subtract "1" from each side of the equation;
x + 1 − 1 = 0 − 1 ;
to get: x = -1
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So, 11x − 1 = 0 ;
Add "1" to each side of the equation;
11x − 1 + 1 = 0 + 1 ;
to get: 11x = 1 ;
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Divide EACH side of the equation by "11"; to isolate "x" on one side of the equation; and to solve for "x" ;
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11x/11 = 1/11; x = 1/11 = 0.0909090909090909.......
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So, yes; there are 2 (two) real solutions.
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