Finding width and length of a garden

Question

Can someone please help me out with this question? I have been at it for hours and I can't wrap my head around this one.

Karen and Kurt's backyard has a width of $20$ meters and a length of $30$ meters. They want to put a rectangular flower garden in the middle of the backyard,leaving a strip of grass of uniform width around all sides of the flower garden. If they want to have $336$ meters of grass, what will be the width and length of the garden?

I had $(20-2x)(30-2x)=336$ for the initial set up, but this didn't work out. I then foiled the two in parenthesis and equaled that out to $336$ then moved the $336$ to the left side so now its:

$$600-100x+4x^2-336=0$$

Ultimately looking like this:

$$ 4x^2-100x+264=0 $$

I then factored out the 4:

$$4(x^2-25x+66) = 0$$ After this part I couldn't find anything with the product of 2 numbers that give me $66$ and $2$ numbers that sum up to $25$ to save my life. I tried other ways and nothing. I am stumped! The answer is:

12 meters wide and 22 meters in length for the garden.

Answer 1

$x^2-25x+66 = (x-22)(x-3) = 0$

I have found these $22$ and $3$ just by quickly checking their divisors (and their suitable sums, products). Infact, there is a method for finding roots (that are possibly hard to guess): For any quadratic equation $ax^2+bx+c=0$, its roots are of the form: $x_{1,2}=\dfrac{-b \pm\sqrt\Delta} {2a}$ where $ \Delta = b^2 - 4ac$.

Answer 2

If $x$ is the width of the strip, your initial setup of $(20-2x)(30-2x)=336$ is fine.

Correction: Like you, I didn’t read closely enough. It’s the grass, not the garden, that is to have an area of $336$ square metres. If you make a sketch, you’ll see that the area of the strip of width $x$ is $$2x(30-2x)+2x(20-2x)+4x^2= 2x(50-4x)+4x^2=100x-4x^2\;,$$ and the correct equation is therefore $100x-4x^2=336$, or $4x^2-100x+336=0$.

Everything in sight is a multiple of $4$, so you can save yourself some work by dividing through by $4$ to get $x^2-25x+84=0$, exactly as you did, albeit to the wrong equation. If you don’t see the factorization $(x-21)(x-4)$, just use the quadratic formula:

$$x=\frac{25\pm\sqrt{25^2-4\cdot84}}2=\frac{25\pm17}2=21\text{ or }4\;.$$

Now it’s clear that the strip can’t be $21$ m wide, so we must have $x=4$, giving the centre plot dimensions of $(20-2\cdot4)\times(30-2\cdot4)=12\times 22$, as stated.