Problem: For x≧0 the smallest value of 6(1+x)4x2+8x+13​ is:
Answer Choices:
A. 1
B. 2
C. 1225​
D. 613​
E. 534​
Solution:
Let y=6(1+x)4x2+8x+13​∴4x2+(8−6y)x+13−6y=0, a quadratic equation in x with discriminant
D=36(y2−4). Since x≧0,D must be non-negative.
∴y≧2, that is, the minimum value of the given fraction is 2 .
Let y=6(1+x)4x2+8x+13​=32(x+1)​+2(x+1)3​ so that y is of the form N+N1​. The minimum value of such an expression occurs when N=N1​, that is, when 32(x+1)​=2(x+1)3​, in other words, when x=21​. With this value of x,y=2.