A Minimum value sum
The minimum value of x 4 − x 2 − 2 4 x + 1 4 5 + x 4 − 2 3 x 2 − 2 x + 1 4 5 can be expressed in the form a b , where a and b are integers, with b is not divisible by the square of any prime. What is the value of a + b ?
The answer is 13.
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1 solution
You must show that equality indeed occurs, to conclude that 1 1 2 is indeed the minimum.
Equality in Triangle inequality occurs when the three points are collinear. So the line joining ( 1 2 , 1 ) and ( 1 , 1 2 ) : y = 1 3 − x must intersect the parabola y = x 2 at some point. Equating we get the equation x 2 = 1 3 − x , and solving we get x = 2 1 ( ± 5 3 − 1 ) . We can check that equality occurs at the positive root.
it is awesome, but i just wonder how many people can actually factorize the equation for first step
everything is perfect! :D except.. you could have mentioned that we can actually use the triangle inequality in this case as the line joining ( 1 , 1 2 ) and ( 1 2 , 1 ) actually cuts the parabola y = x 2 because, if that were not the case, we would have to think differently... i just think that pointing out the above fact completes the whole solution :D
That is just awesome!!
Instead of geometry, use, √a+√b≥(ab)^1/4 Where a,b are compared expressions with the question.
At minimum point, √a+√b=(ab)^1/4 Or,a=b, Now simply solve the equation. And use the value of x in (ab)^1/4. This is the requirement
I am the 244 the solver
there is another solution,we can derivate whole function and equal to 0.Those are the candidates for minimum.After manipulating we get 4x^7-48x^6-102x^5+1219x^4+528x^3-7165x^2-314x+1885=0, it's obvious that this polynome is diveded by x^2+x-13,so we get (x^2+x-13)(4x^5-52x^4+2x^3+541x^2-145)=0,now we find all roots annd get values -4.14,3.14,-2.84,-0.54,0.51,3.84,12.0226 and we calculate that minimum is about 3.14 which is positive solution of x^2+x-13=0,exactly x=(-1+√53)/2. Now we replace this value in our function and we get after few calculations that minimum value is 11√2...
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You're derivative manipulation is incorrect, which explains the plethora of extraneous solutions. When you set the derivative equal to zero, by the first derivative test, there are only 3 solutions possible, as the polynomial is cubic. The only three solutions are 3.140005, -2.84455, and -0.538474. Since f(x) changes from negative to positive when x = 3.14, then f(3.14) is a minimum, and f(3.14) yields 11 root 2. The bottom line is, the manipulation you created, which yielded your septic equation, is absolutely incorrect (hence the 4 extraneous solutions.) Such an algebraic manipulation, is not even possible. So yeah, sorry!
The above expression can be factorized into:
( x 2 − 1 ) 2 + ( x − 1 2 ) 2 + ( x 2 − 1 2 ) 2 + ( x − 1 ) 2
Now, we have to think geometrically :)
By observation, the above factorized expression is a form of Distance Formula on the Cartesian Plane.
Hence, we get the points on the Cartesian Plane as ( x 2 , x ) , ( 1 , 1 2 ) and ( 1 2 , 1 )
Let us join these points to see the magic :)
After joining the points, we get a Triangle.
Now, let the sides of the triangle be a , b and c where c is the side representing the distance between the points ( 1 , 1 2 ) and ( 1 2 , 1 ) .
By the Triangle Inequality in the above triangle formed, we get:
a + b ≥ c
c = 1 1 2 (By using the Distance Formula between the points ( 1 , 1 2 ) and ( 1 2 , 1 )
Hence, the minimum value of our original expression is 1 1 2 .