Question 1208228: If a + b = 1 and a⁴ + b⁴ = 17.
Find a,b
Found 2 solutions by mananth, ikleyn:
Answer by mananth(16946)   (Show Source):
You can put this solution on YOUR website! If a + b = 1 and a⁴ + b⁴ = 17.
If a + b = 1 and a⁴ + b⁴ = 17.
(a+b)=1
square the equation
(a+b)^2=1
a^2+2ab+b^2=1
a^2+b^2=1-2ab
Now
a^4+b^4=17 (given)
use identity
(a^2+b^2)^2 -2a^2b^2=17
(1-2ab)^2 -2(ab)^2=17
1-4ab+4a^2b^- 2(ab)^2=17
-4ab+2(ab)^2=16
2a^2b^2 -4ab-16=0
2(ab)^2-4ab-16=0
compare with ax^2+bx+c=0
here a = 2 b=-4 c= -16
Use formula method
b^2-4ac = 16 +128=144
ab =( 4+12)/4 OR 4 -12/4
ab = 4 or -2
Now
If ab = 4
a= 4/b
a+b=1 (given)
substitute a
4/b +b=1
Multiply by b
4+b^2=b
b^2-b+4 =0
the discriminant is <0 no real roots
When ab =-2
a+b=1
ab =-2
b= -2/a
a-2/a = 1
Multiply by a
a^2-2=a
a^2-a-2=0
Use quadratic formula
we get
a=2 or -1
Plug a in a+b =1 find b for both values of a
Answer by ikleyn(52786)  (Show Source):
You can put this solution on YOUR website! .
Two solutions can be guessed mentally and instantly:
(a,b) = (2,-1) and (a,b) = (-1,2). (1)
It can be proved mentally and visually.
The curve x^4 + y^4 = 17 is similar to the circle x^2 + y^2 = 17.
It is not a circle, of course, but is a similar convex shape.
The line x + y = 1 is the straight line that cuts this curve x^4 + y^4 = 17.
It is clear, that there are two intersection points, and can not be more than 2 intersection points.
Thus, the listed solutions (1) is the full set of real solutions.
If you want to get an algebra solution, then express
a = 1 - b
from the first equation and substitute into the second equation.
You will get a polynomial equation P(b) = 17 for "b" of degree 4.
Two roots are b = -1 and b = 2.
Hence, two linear divisors are (b+1) and (b-2).
Divide the polynomial P(b)-17 by the product (b+1)*(b-2).
You will get the quotient as a quadratic polynomial, which has no real roots.
From it, you will conclude that the listed solutions (1) are the only solutions, and there are no other solutions.
It is how the problem looks like and everything behind and around it.
|
|
|
