a +b = c a x b = c what the value of a, b, and c with value of a &b not 2 & 0

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a +b = c a x b = c what the value of a, b, and c with value of a &b not 2 & 0

Precalculus
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If you knew what c was, then solving the system\[a+b=c\]\[ab=c\]Is the same as solving the quadratic equation:\[x^2-cx+c=0\]where the solutions to the quadratic are a and b. Using the quadratic formula yields:\[x=\frac{-c\pm \sqrt{c^2-4c}}{2}\] Now we probably want these solutions to be real, so we need to make sure:\[c^2-4c\ge 0\]If c is positive, and not 0, then we divide by c to get:\[c-4\ge 0\Longrightarrow c\ge 4\] So pick any number for c such that c is greater than 4. Say 10. Then:\[a=\frac{-10+\sqrt{60}}{2}\]\[b=\frac{-10-\sqrt{60}}{2}\]does the job.
Note there is not one answer to this question. You can pick infinitely many values of c, which will produce a and b accordingly.
whoops typo, should be:\[a=\frac{10+\sqrt{60}}{2}\]\[b=\frac{10-\sqrt{60}}{2}\]

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Other answers:

ab = a+b ab - a = b a(b-1) = b a = b/(b-1), with b not 1 just plug b E real but not for b=0,1,and 2 u will get values a and get order pairs : (a1,b1),(a2,b2),.... so, infinitely solutions

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