Math Is Fun Forum

(x-A)*(x-B)*(x-C)*(x-D)*(x-E) = 0

x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] - A*B*C*D*E = 0
or
x^5 - (x^4)*(A + B + C + D + E) + (x^3)*(A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E) - (x^2)*(A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E) + x*(A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E] = A*B*C*D*E

We have:
x^5 - 10x^4 + ax^3 + bx^2 + cx = 320

By comparing, we get
A + B + C + D + E = 10
A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = a
A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = b
A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = c
A*B*C*D*E = 320

A + B + C + D + E = 10

In general, Sum = [2*m + (n-1)*r]*n/2
where
m is the first term
r is the common difference
n the number of terms

[2*A + (5-1)*r]*5/2 = 10
A + 2*r = 2
r = (2 - A)/2

For A = -4 , r = 3
Therefore
A = -4
B = -1
C =  2
D = 5
E = 8

Their sum is 10
And their product is 320

a = A*B + A*C + A*D + B*C + B*D + C*D + A*E + B*E + C*E + D*E = -5
b = A*B*C + A*B*D + A*C*D + B*C*D + A*B*E + A*C*E + A*D*E + B*C*E + B*D*E + C*D*E = 190
c = A*B*C*D + A*B*C*E + A*B*D*E + A*C*D*E + B*C*D*E = -136

Therefore:
a + b + c = 49

Kerim

I did a mistake (sorry, a mistyping, not about the asnswer); will someone correct it?

This exercise has a solution if the right side of the equation is 32 instead of 320.

Well done, I couldn't do it better.

Good remark.
In general, a problem may have more than one solution (even more than two as it is the case here).
In real life, one usually chooses the solution which seems be the best one for his application.
For example, in this exercise, we use to see the base of a rectangle be longer than its height. So, our preference is likely to say that h=4.5 cm and b=12 cm. But this is valid as long no one complains (mainly a math teacher at one's school).   

At the start we had indeed two simultaneous equations to solve. But, from them, we got a quadratic one to be solved.

I mean, let us consider the following two simultaneous equations.
3x + 2y - 7 = 0 [eq. 1]
x + 4y - 9 = 0 [eq. 2]

Multiplying [eq. 2] by 3
3x + 12y - 27 = 0 [eq. 3]

From [eq. 3] - [eq. 1], we get
0x - 10y - 20 = 0
10y - 20 = 0 [eq. 4]

Here, we derived a linear equation from the two simultaneous ones instead of a quadratic equation.

In both cases, we had two simultaneous equations to solve

Anyway, perhaps someone knows better what solving simultaneous equations does mean.

Well done.

if Ax^2+Bx+C=0
x=[-B±√(B^2-4AC)]/2A

Here:
2b^2 - 33b + 108 = 0
A=2
B=-33
C=108

Therefore, if b=12 cm, h=54/12=4.5 cm
And, if b=4.5 cm, h=54/4.5=12 cm

[a*a*a*a... m times / a*a*a*a... n times] = a^m/a^n = a^(m-n)
Similary, for any value of x we can write:
1 = a^x/a^x = a^(x-x) = a^0

You are right.

I mean, see what happened to Galileo for example. He chose to live hard days for presenting openly his discoveries which were not supposed to be heard by the world's multitudes of his time.

From the original equation:
[1] x-1801+sqrt[(y-1860)*(x-1801)] = 2*sqrt(x-1801) - 1
[2] sqrt[(y-1860)*(x-1801)] = -(x-1801)+ 2*sqrt(x-1801) - 1
[3] sqrt[(y-1860)*(x-1801)] = -[sqrt(x-1801) - 1]^2

Also from the original equation:
x-1801 > 0
x > 1801

From the equation [3] above:
The right side is always negative.
And for the sqrt on left side to exist, the right side should be 0.
[4] [sqrt(x-1801) - 1] = 0
[5] sqrt(x-1801) = 1
[5] (x-1801) = 1
[6] x = 1801 + 1
[7] x = 1802

Also the left side of [3] should be zero.
[8] sqrt[(y-1860)*(x-1801)] = 0
[9] (y-1860)*(x-1801) = 0
By replacing x = 1802
[10] (y-1860)* 1 = 0
[11] (y-1860) = 0
[12] y = 1860

Therefore, there is one solution only:
x = 1802
y = 1860

Indeed, this is the first step in designing every part of a new application.

The main, if not crucial, role of the human scientific brain is to find out the formula, the function or the equation which can emulate/reflect, as possible, a real problem that needs to be solved.

Therefore, the purpose of the various math's exercises is to gain, also as possible, the logical reasoning on how to get the required/needed results from analyzing a formula/function of interest or solving a well-defined equation(s). Fortunately, there are also many ready-made tools to assist someone in doing this to save time (after he learns how to use them).

The calculus way is based on the knowledge of the derivatives of functions.