The average of two numbers is 41.125, and their product is 1683. What are the numbers?

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Answer 1 · 2017-02-23T06:29:08

The two numbers are $38.25$ $38.25$ and $44$ $44$

Let the numbers be $a$ $a$ and $b$ $b$ .

As their average is $\frac{a + b}{2}$ $(a+b)/2$ , we have $\frac{a + b}{2} = 41.125$ $(a+b)/2=41.125$

or $a + b = 41.125 \times 2 = 82.25$ $a+b=41.125xx2=82.25$

or $a = 82.25 - b$ $a=82.25-b$ i.e. the numbers are $(82.25 - b)$ $(82.25-b)$ and $b$ $b$

As the product of numbers is $1683$ $1683$ , therefore

$b (82.25 - b) = 1683$ $b(82.25-b)=1683$

or $82.25 b - b^{2} = 1683$ $82.25b-b^2=1683$

or $329 b - 4 b^{2} = 6732$ $329b-4b^2=6732$ - multiplying each term by $4$ $4$

i.e. $4 b^{2} - 329 b + 6732 = 0$ $4b^2-329b+6732=0$

and using quadratic formula $b = \frac{329 \pm \sqrt{329^{2} - 4 \times 4 \times 6732}}{8}$ $b=(329+-sqrt(329^2-4xx4xx6732))/8$

= $\frac{329 \pm \sqrt{108241 - 107712}}{8} = \frac{329 \pm \sqrt{529}}{8}$ $(329+-sqrt(108241-107712))/8=(329+-sqrt529)/8$

= $\frac{329 \pm 23}{8}$ $(329+-23)/8$

i.e. $b = \frac{352}{8} = 44$ $b=352/8=44$ or $b = \frac{306}{8} = \frac{153}{4} = 38.25$ $b=306/8=153/4=38.25$

ans $a = 82.25 - 44 = 38.25$ $a=82.25-44=38.25$ or $a = 82.25 - 38.25 = 44$ $a=82.25-38.25=44$

Hence the two numbers are $38.25$ $38.25$ and $44$ $44$