a+b=8
ab+c+d=23
ad+bc=28
cd=12
solve in real numbers the system of equation
these points are gifts for those who solve it
Answers
a b + c + d = 23 -- equation 2
a d + b c = 28 -- equation 3
c d = 12 --- equation 4
From equation 1 and 4, we get
b = (8 - a) -- equation 5
c = 12/d -- eq 6
Substitute these two values in equation 3
a d + (8 - a) (12/d) = 28
12 a + 28 d - a d^2 - 96 = 0
a = 4 (24 - 7 d) / (12 - d^2) --- eq 7
Substitute this in equation 5 to get,
b = 4 d (7 - 2 d) / ( 12 - d^2) -- eq 8
Substitute values from equations 7, 8 and 4 in equation 2. We get
16d (24 - 7d) (7 - 2d) / (12 - d^2)^2 + 12/d + d = 23
16d^2 (168 - 97 d + 14 d^2) + (12 - d^2)^2 (12 -23 d + d^2) = 0
Simplifying further,
P(d) = d^6 - 23 d^5 + 212 d^4 - 1000 d^3 + 2544 d^2 - 3312 d + 1728 = 0 -- eq 9
There are six real roots for this equation. They are all positive - can be found found from Descarte's rule for zeroes.
Roots are factors of 12^3 or 1728 ie., from the list of 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, etc.
Trying 2, 3, 4, and 6. They are satisfied. Now, we can write the 6th degree polynomial in d as follows. here x is not known yet.
P(d) = (d - 2) (d - 3) (d - 4) (d - 6) [d^2 - x d + 1728/(2*3*4*6) ] = 0
Comparing the coefficient of d^3, -23 = -x - 10 -5 => x = 8
factors of (x^2 - 8 d + 1728/144) are (d - 2) and (d - 6)
So the roots of P(d) are d = 2, 3, 4, and 6. Here 2 and 6 occur twice.
Now for solving the given system of equations. Given equations are symmetric in a and b. They are also symmetric in c and d. Hence there will be solutions with interchange of values of a and b, as well as c and d.
Solution 1:
d = 2, c = 6 , a = (96-56)/(12-4) = 5, b = 8 - 5 = 3
Solution 2
d = 3, c = 4, a = (96 - 84)/(12 - 9) = 4, b = 8 - 4 = 4
Solution 3
d = 4, c = 3, a = (96 - 112)/ (12 - 16) = 4 , b = 4
Solution 4
d = 6, c = 2, a = 3 , b = 8 - 3 = 5
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solutions are : (5, 3, 6, 2) , (4, 4, 3, 4), (4, 4, 4, 3) and (3, 5, 2, 6)
Answer:
a + b = 8 -- equation 1
a b + c + d = 23 -- equation 2
a d + b c = 28 -- equation 3
c d = 12 --- equation 4
From equation 1 and 4, we get
b = (8 - a) -- equation 5
c = 12/d -- eq 6
Substitute these two values in equation 3
a d + (8 - a) (12/d) = 28
12 a + 28 d - a d^2 - 96 = 0
a = 4 (24 - 7 d) / (12 - d^2) --- eq 7
Substitute this in equation 5 to get,
b = 4 d (7 - 2 d) / ( 12 - d^2) -- eq 8
Substitute values from equations 7, 8 and 4 in equation 2. We get
16d (24 - 7d) (7 - 2d) / (12 - d^2)^2 + 12/d + d = 23
16d^2 (168 - 97 d + 14 d^2) + (12 - d^2)^2 (12 -23 d + d^2) = 0
Simplifying further,
P(d) = d^6 - 23 d^5 + 212 d^4 - 1000 d^3 + 2544 d^2 - 3312 d + 1728 = 0 -- eq 9
There are six real roots for this equation. They are all positive - can be found found from Descarte's rule for zeroes.
Roots are factors of 12^3 or 1728 ie., from the list of 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, etc.
Trying 2, 3, 4, and 6. They are satisfied. Now, we can write the 6th degree polynomial in d as follows. here x is not known yet.
P(d) = (d - 2) (d - 3) (d - 4) (d - 6) [d^2 - x d + 1728/(2*3*4*6) ] = 0
Comparing the coefficient of d^3, -23 = -x - 10 -5 => x = 8
factors of (x^2 - 8 d + 1728/144) are (d - 2) and (d - 6)
So the roots of P(d) are d = 2, 3, 4, and 6. Here 2 and 6 occur twice.
Now for solving the given system of equations. Given equations are symmetric in a and b. They are also symmetric in c and d. Hence there will be solutions with interchange of values of a and b, as well as c and d.
Solution 1:
d = 2, c = 6 , a = (96-56)/(12-4) = 5, b = 8 - 5 = 3
Solution 2
d = 3, c = 4, a = (96 - 84)/(12 - 9) = 4, b = 8 - 4 = 4
Solution 3
d = 4, c = 3, a = (96 - 112)/ (12 - 16) = 4 , b = 4
Solution 4
d = 6, c = 2, a = 3 , b = 8 - 3 = 5
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solutions are : (5, 3, 6, 2) , (4, 4, 3, 4), (4, 4, 4, 3) and (3, 5, 2, 6)
Step-by-step explanation: