What two real numbers have a sum of 8 and a product of 2?

Answer 1

x + y = 8

x * y = 2

First let's solve the first equation for x in terms of y:

x + y = 8

x = 8 - y

Now let's plug that into our second equation and simplify it:

x * y = 2

(8 - y) * y = 2

8y - y2 = 2

-y2 + 8y - 2 = 0

Now the always fun quadratic equation to find y:

(Since I don't know how the system will respond to a "plus or minus" sign, I'll use +/- instead)

-b +/- sqrt(b2 - 4*a*c)

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2*a

...and plugging in our values (a = -1, b = 8, c = -2)

-8 +/- sqrt(82 - 4(-1)(-2))

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2(-1)

-8 +/- sqrt(64 - 8)

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-2

-8 +/- sqrt(56)

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-2

We'll use a calculator to figure out that sqrt(56) = 2*sqrt(14)

-8 +/- 2*sqrt(14)

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-2

Please excuse the formatting...

-8/-2 +/- 2*sqrt(14)/(-2)

4 +/- (-sqrt(14))

4 -/+ sqrt(14)

Note the shift in the +/- to -/+ when we multiply out that -1

y = 4 -/+ sqrt(14)

Now that we know y, let's go back to our very first equation and plug in the value.

x = 8 - y

x = 8 - (4-/+sqrt(14))

Now we have two possible values of y, which change the possible value of x:

When y = 4 - sqrt(14),

then x = 4 + sqrt(14)

When y = 4 + sqrt(14),

then x = 4 - sqrt(14)

In conclusion, the two real numbers which have a sum of eight and a product of two are:

x = 4 - sqrt(14), y = 4 + sqrt(14)

x = 4 + sqrt(14), y = 4 - sqrt(14)