Math CentralQuandaries & Queries


Question from marj, a student:

find two real numbers whose sum is 22 and whose product is 125.

Hi Marj,

Suppose the two numbers are $x$ and $y.$ Their sum is 22 so

\[x + y = 22.\]

Their product is 125 and this gives you another equation. Solve the first equation for y, substitute into the second equation and simplify. Solve the resulting quadratic.

If you have a hard time believing the result use some computer software or a graphing calculator to plot the quadratic.



Eliminate one variable: $x+y = 22 \rightarrow y = 22-x$

Plug in: $125 = x(22-x);$

Tidy up: $x^2 - 22x + 125 = 0$

Now, the quadratic formula requires you to take the square root of $b^2 - 4ac$ where (here)$ a=1, b=-22, c = 125$

Can you do this? (If not, there is no solution.)

Good Hunting!


Let x an y be the two real numbers with the desired properties
Then $x+y=22$ (sum is $22$) so $y = 22-x (*)$
And $xy=125$ (product is $125$) $(**)$
Substituting equation * into ** gives: $ x(22-x)=125$
Expanding gives: $22x - x^2 = 125$
Rearranging gives $x^2 - 22x + 125 = 0$ (if you knew the relationship between quadratic equations and their roots, then you could have jumped directly to this equation from the question)
Using the quadratic formula gives:

\[x = \frac{22 \pm \sqrt{484 - 500}}{2}\]
These generates two complex number roots.
Hence, there are no REAL numbers satisfying the desired properties!


About Math Central


Math Central is supported by the University of Regina and the Imperial Oil Foundation.
Quandaries & Queries page Home page University of Regina