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Can you create a quadratic equation with the solution being -5? - Answers
Yes, this can be done this way:(x-(-5))(x+a)=0, where a is any (positive or negative) numberThis simplifies to: (x+5)(x+a) = 0or written as quadratic: x2+(a+5)x+5a = 0Solutions: x1 = -5 and x2 = -a,so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )and there will be only one solution: x1,2 = -5
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Can you create a quadratic equation with the solution being -5? - Answers
Yes, this can be done this way:(x-(-5))(x+a)=0, where a is any (positive or negative) numberThis simplifies to: (x+5)(x+a) = 0or written as quadratic: x2+(a+5)x+5a = 0Solutions: x1 = -5 and x2 = -a,so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )and there will be only one solution: x1,2 = -5
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Can you create a quadratic equation with the solution being -5? - Answers
Yes, this can be done this way:(x-(-5))(x+a)=0, where a is any (positive or negative) numberThis simplifies to: (x+5)(x+a) = 0or written as quadratic: x2+(a+5)x+5a = 0Solutions: x1 = -5 and x2 = -a,so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )and there will be only one solution: x1,2 = -5
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- og:descriptionYes, this can be done this way:(x-(-5))(x+a)=0, where a is any (positive or negative) numberThis simplifies to: (x+5)(x+a) = 0or written as quadratic: x2+(a+5)x+5a = 0Solutions: x1 = -5 and x2 = -a,so if a=5 the equation shortens to (x+5)2=0 ( or x2+10x+25=0 )and there will be only one solution: x1,2 = -5
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