What is the smallest possible real value of \(x^2+8x\)?

 Apr 24, 2019

Definitely not positive integers... because of the +8x

Negatives can't be too small because of the \(x^2\) will turn it into a positive integer.


Because I'm am not the best at math, I will just use a guess and check to solve.


1. We are trying to find the peak at which the value will be on the verge of becoming bigger.


    Substituting -3 for x we will get : -15


    Substituting -4 for x we will get : -16


    Substituting -5 for x we will get : -15


So the answer SHOULD be -16? I'm not sure someone check this.

 Apr 24, 2019

\(\text{For this problem the simplest route is just completing the square}\\ y=x^2+8x = \\ x^2 + 8x + 16 - 16 = \\ (x+4)^2 - 16\\ \text{This is a minimum at }x=-4 \text{ and the minimum is }y=-16\)

 Apr 24, 2019

So since we are trying to find the value of the expression, does that mean that the Y value is the answer?

CalculatorUser  Apr 24, 2019

your answer was correct, and guessing has it's place in solving problems,

but there was a concrete way to solve this one.

Rom  Apr 25, 2019

Hi CalculatorUser,

A big thanks to you and Rom.


To answer your question... Yes it does.


y= x^2+8x     is a concave up parabola.   (so it will have a minimum value)


So the 2 roots are at x=0 and  x=-8

The axis of symmetry is half way between them  x=(0+-8)/2 = -4

The minimum will be the y value when x=-4


y=-4(-4+8) = -4*4 = -16



sketch what I have done so you can see what I am talking about. 

Melody  Apr 25, 2019

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