Find the sum of integer values that satisfy x2 + 6x - 7<0.
Answers
2x + 6x - 7 < 0
= 8x - 7 < 0 (simplify)
= 8x < 7
x = 7/8
x = .875 (in decimal form)
Variable "x" can also equal anything lower in value to than in relation of 7/8 (.875)
Step-by-step explanation:
Given :-
x^2 + 6x - 7<0.
To find:-
Find the sum of integer values that satisfy
x^2 + 6x - 7<0.
Solution:-
Given in-equation is x^2 + 6x - 7<0.
=> x^2 - x + 7x -7 < 0
=>[ x(x-1) +7(x-1) ]<0
=> (x-1)(x+7)<0
Product of two things is less than zero i.e. one of them must be positive and other must be Negative.
Also since (X-1 ) is smaller than (x+7) therefore (X-1) must be negative
=> x -1 <0 or x+7 > 0
=> x -1+1+< 0+1 or x+7-7>0-7
=> x<1 or x>-7
=> -7< x< 1
The possible values of x are -6,-5,-4,-3-,2,-1,0.
Their sum =
(-6)+(-5)+(-4)+(-3)+(-2)+(-1)+0
= -21+0
= -21
Their sum = -21
Answer :-
The sum of integer values that satisfy
x^2 + 6x - 7<0 is -21
Used Concept:-
Product of two things is less than zero i.e. one of them must be positive and other must be Negative.