Vtyjkyuk

It is given that area of rectangle = its perimeter and the diagonal is 3√5. We are to find it's area by use of quadratic equations. The problem is that I'm getting a 4 degree equation when I substitute for a in the second equation. How do I find the sides of rectangle ?

Let, length$=a$ and weidth $=b$
So,
$$ab=2(a+b)..............(1)$$
$$textand,$$
$$a^2+b^2=(3 sqrt(5))^2...............(2)$$
From (2) we get,

beginarrayll
& a^2+b^2 &=(3 sqrt(5))^2\
& implies a^2+b^2 +2ab &=(3 sqrt(5))^2 +2ab \
& implies (a+b)^2 &=45+4(a+b) \
& implies (a+b)^2 -4(a+b)-45 &=0 \
&Let,&\
& & a+b=p\
& implies p^2 -4p-45 &=0 \
& implies (p-9)(p+5)&=0 \
& p=9,-5 &[textHere, p=-5 is not possible, because the sum of two lengths can't be negative]\
&Hence,&\
& a+b &=9 \
& implies a &=9-b..........(3)
endarray

From equation (1) we get now,

beginarrayll
& (9-b)b &=2(9-b+b)\
& implies b^2-9b+18 &=0 \
& implies (b-6)(b-3) &=0 \
& implies b &=3 text or, 6
endarray

From (3) we get,
$$ a=6,text if b=6$$
$$ a=3,text if b=3$$

From
$$(3sqrt5)^2=a^2+b^2=(a+b)^2-2ab=frac14(ab)^2-2ab$$
$$iff (ab)^2-8ab-180=0iff(ab-18)(ab+10)=0$$
we find $ab=18$ or $-10$. Hence $a+b=9$. Now consider $a$ and $b$ as the solutions of
$$x^2-9x+18=0iff(x-3)(x-6)=0.$$

Let's denote by $a,b$ the sides of the rectangle. Since the area is equal to the perimeter we have that

$$ab=2(a+b)$$ and using the information about the diagonal it is

$$sqrta^2+b^2=3sqrt5.$$ Squaring we get

$$a^2+b^2=45.$$

Thus, one has to solve the system

begincases
2a+2b &=ab\
a^2+b^2&=45
endcases

From the first equation we get $b=dfrac2aa-2$ and subtituting in the second it is

$$a^2+dfrac4a^2(a-2)^2=45.$$ We get

$$a^4-4a^3-37a^2+180a-180=0.$$

The integer solutions of this equation are divisors of $180.$ Since $a$ is the side of a rectangle we don't consider negative values. We see that $1$ and $2$ are not solutions but $3$ is. So we have

$$(a-3)(a^3-a^2-40a+60)=a^4-4a^3-37a^2+180a-180=0.$$ Now we work with $$a^3-a^2-40a+60$$ to find more possible solutions. We see that $6$ is a solution. And

$$a^3-a^2-40a+60=(a-6)(a^2+5a-10).$$ Thus

$$a^4-4a^3-37a^2+180a-180=(a-3)(a-6)(a^2+5a-10)=0.$$

So $$(a,b)=(3,6):textand: (a,b)=(6,3)$$ are integer solutions of the problem.

Now we consider the equation $a^2+5a-10=0.$ It has a positive solution which is $$dfracsqrt65-52.$$ Since

$$dfracsqrt65-5dfracsqrt65-52-2<0$$ this doesn't give a solution to the problem.