Week 9 – Pre-Calc11 – Owen's Blog

This week in pre-calculus I finally figured out how to solve quadratic problems with two variables!

Let’s say we have two number with a difference of 18 and the sum of their squares is a minimum. How do we find the numbers?

First off, we know $x-y=18$

We can also say $x=18+y$

And knowing that the sum of their squares is a minimum, we can say that $S=x^2 + y^2$

But since $x=18+y$ we can write out the equation as $S=(18+y)^2 +y^2$

Now looking at this we can say $(18+y)(18+y)+y$

Expand this out and we end up with $2y^2 + 36y + 324$

As you may recognize, we have reached the general form!

Now we complete the square to bring this to standard/vertex form which will be $2(y+9)^2 + 162$

Having this we now know out vertex which will be (-9,162)

Now from this vertex, we have our y value which is -9 and our minimum. To find x, all we do is plug 9 back into the original equation

$x+9= 18$

and here we can say x=-9

Thanks for checking this out.

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