The sum of [x] for all real numbers x satisfying the equation 16+15...
The sum of [x] for all real numbers x satisfying the equation 16+15x+15x2=[x]3 is:
Correct answer
33
Step-by-step solution
[x]3=15x2+15x+16[x]∈(x−1,x][x]3∈((x−1)3,x3]⇒15x2+15x+16∈((x−1)3,x3]x3≥15x2+15x+16⇒x3−15x2−15x−16≥0(x−16)(x2+x+1)≥0⇒x≥16
Also, 15x2+15x+16>x3−3x2+3x−1⇒x3−18x2−12x−17<0
Let f(x)=x3−18x2−12x−17f(x)=3x2−36x−12 f(α)<0⇒f(x) has only one real root ⇒f(18)=−233f(19)=116⇒f(α)=0⇒α∈(18,19)
For this equation
x∈[16,19)⇒[x]=16,17,18
(i) [x]=16⇒⇒15x2+15x+16=163(x−16)(x+17)=0x=16 satisfies
(ii) [x]=17⇒15x2+15x+16=173⇒x=30−15±152−4.16.15⇒[x]=17 satisfies
(iii) [x]=18⇒x>19 and x<−20⇒ No received value of x⇒ Sum of [x]=16+17=33.