Quadratic equation biting its own tail

a+b=c (1) similarly for the other equation c+d=a (2). Add (1) & (2) to get b+d=0 which gives b=d=0 as they must be non-negative. using these in (1) & (2) we get a=c. It is easy to verify that the conditions a=c; b=d=0 imply the data.

In the general quadratic equation ax^2 + bx + c = 0, vieta’s formula states that the sum of the roots is -b/a and the product is c/a.

And so we shall apply vieta’s formula. So for x^2 - cx + d = 0 with roots a and b, we have that

a+b=c ---EQ1

ab=d ---EQ2

Similarly for x^2 - ax + b = 0 with roots c and d

c+d=a --- EQ3

cd=b --- EQ4

We notice that EQ 1 is similar to EQ3. Therefore, we try to add them together and see what happens. So we get a+b+c+d=a+c,which simplfies to b = -d, but then we notice that b and d must be positive and therefore b=d=0 since 0 <= a, b, c, d <= 100 and EQ1 and EQ3 now reduce to a=c

So the solution set of quadruples is (n,0,n,0) for every n in the range 0 <= n<= 100.

i.e. are 101 quadruples (since we must include (0,0,0,0)).

we have two equations; x^2-cx+d=0; (1) x^2-ax+b=0; (2) as we know that the coefficient of the term with order 1, equals -1 times sum of the roots, and constant term equals the product of the roots, hence; from eqn (1); -c= -(a+b); d= a b; and from eqn(2); -a= -(c+d); b=c d; so we have; c-b=c+d; -b=d; but as given, that 0<=(a,b,c,d)<=100; hence b=d=0; and a=c; (check above equations and see) so the quadruples are (a,0,a,0), total 101 (don't forget to include zero)

By the fundamental theorem of algebra we know $x^2-cx+d$ , $x^2 - ax + b$ two roots. Noting that the leading coefficients are both $1$ , it follows that we have no constant factors.

Given $a,b$ are roots of $x^2-cx+d=0$ , we then know they are the only roots of $x^2-cx+d=0$ ; similarly, we know $c,d$ are the only roots of $x^2-ax+b=0$ . Thus by the factor theorem we may rewrite $x^2-cx+d=(x-a)(x-b)$ and $x^2-ax+b=(x-c)(x-d)$ .

Expanding both products out, we have

$\begin{aligned}x^2-cx+d&=x^2-ax-bx+ab\\&=x^2-(a+b)x+ab\end{aligned}$ $\begin{aligned}x^2-ax+b&=x^2-cx-dx+cd\\&=x^2-(c+d)x+cd\end{aligned}$ .

Since polynomial forms are unique, we may equate coefficients to yield

$c=a+b\\d=ab\\a=c+d\\b=cd$

One trivial apparent solution is $(0,0,0,0)$ .

Notice that if $b$ is non-zero, then necessarily $a,c,d$ are non-zero and furthermore $a=1/c$ ; since $a,c$ are positive integers, both must then be $1$ (the only positive integer who is its own multiplicative inverse). If $a,c$ are $1$ , however, then $b,d$ must be $0$ so that $a+b=c$ and $c+d=a$ -- a contradiction. Thus $b$ must be zero.

It follows readily that $d$ must then be zero as well. Thus $c=a$ , and so any given $a$ , any other variable is determined. Noting that in $[0...100]$ there are $101$ possible choices for $a$ , the number of quadruples, then, is merely (by the fundamental counting principle) $101\times1\times1\times1=101$ .

Vieta's Formulas say that if the equation ax^2 + bx + c = 0 has roots r and s, then r+s = -b/a and rs = c. This can be proven by expanding a(x-r)(x-s) = 0.

Using this, we obtain that a+b = c. So, a = c-b. Also, by Vieta's, c+d = a = c-b. That means that d = -b.

Vieta's also says that ab = d. If b is nonzero, that means that a = d/b = d/(-d) = -1. But the problem says that a is at least 0, so b must be 0.

Therefore, d = 0. That means that a = c. Since 0 <= a,c <= 100, the number of possible quadruples (a, b, c, d) must be 101.

As 'a' and 'b' are the roots of the first quadratic expression The product of roots is given by

ab= d. (By formula for product of roots of a quadratic expression)

The product of roots of the second quadratic expression is given by

cd= b

But d=ab

Hence cab=b

Now there are 2 cases .b is either 0 or non zero

Case1

If b not zero. Implies

Ca=1 ( as cab= b)

Since c and a are integers thus there is only 1 possibility c=a=1

If the above case were true then consider the sum of roots of the first quadratic expression with roots a and b.

a+b=c. (From formula of sum of roots in a quadratic expression)

Since a=c =1

Implies b=0

But this arose from a case where we assumed b to be non zero .hence our assumption was wrong and b is indeed 0

But

ab=d

Hence d is also zero

As a+b=c

And b is zero

a=c

So we need to find in how many ways 'a' can be equal to 'c' in the range of 0 to 100(as b and d are always equal to zero the solution set just contains values that a and c can take)

As 'a' and 'c' are equal

we need to just find the number of integers between 0 and 100 Which is 101

Using the property of roots of quadratic equations, we get 2 sets of 2 equations:

a+b=c (1) ab=d (2)

c+d=a (3) cd=b (4)

Adding (1) and (3) we get : a+b+c+d=a+c So b+d=0 or b = - d

Substitute this in (2) to get : -ad=d

Here we have two cases: d=0 or d!=0

If we take d!=0, then it means a=-1 , b=0, c=-1 and also d=0 (contradiction)

So we take d=0 and b=0 From (1) and (2) we get a=c and c=a

This is a consistent set of solutions

The set of solutions will be (x,0,x,0) where 0<=x<=100 So we get 101 solutions

We notice that a+b=c and a b=d, since a and b are roots of the equation x^2-c x+d=0. For the same reason c+d=a, c d=b. Solving the equations: a+b=c, a b=d, c+d=a, c*d=b for integers between 0 and 100 we find that either b and d are both zero and a=c or all a, b, c and d=0. That means 100 + 1=101 cases.

From Vieta we have a+b=c, ab=d, cd=b, c+d=a. Substituting a+b for c and ab for d into the last equation, we get b(a+1)=0. Since a must be nonnegative, b=0, which implies d=0. Then we have a=c. Since a can range from 0 to 100, this gives 101 solutions.

By the Vieta's formula, the given conditions can be rewritten as the following system of equations: $\begin{cases} a+b=c\\ ab=d\\ c+d=a\\ cd=b \end{cases}$

Hence, this implies that $a + b + c + d = a + c$ , or that $b + d = 0$ . Since the terms are all non-negative, $b=d=0$ .

Because $b=0$ , from the first two equations we get $a=c$ and $d=0$ . Note that the quadruples $(a,0,a,0)$ satisfy the given conditions, for all $a$ from $0$ to $100.$ So the system has $101$ integer solutions in the given range.