If we would like to satisfy the condition:
x^2−11x−210 < 0,
we need to find the roots of
x^2−11x−210=0.
By factoring, we have
(x-21)(x+10)=0
The leading coefficient (of x^2) is positive, so the curve is concave up.
=> x<0 between -10 and 21.
To find the value of a, we convert the absolute value inequality into two ordinary inequalities as:
x-8<a and 8-x <a
but we need to make sure that both inequalities are satisfied.
Can you take it from here?
Find the maximum value of a such that the following statement is true:
If ∣x−8∣<a, then x^2−11x−210<0.
4 answers
Thanks, but how do I find the value of x?
You probably would like to find the value of a.
We established that to satisfy the quadratic inequality
x^2−11x−210 < 0
-10<x<21
Now substitute this interval into each of the two linear inequalities:
x-8<a and 8-x <a
(-10)-8<a => a>-18
8-(-10)<a => a> 18
(21)-8<a => a>13
8-(21)<a => a>-13
To satisfy all four inequalities, we need a>18.
Finally check at least a point between -10 and 21, say 0.
0-8<18, 8-0<18
so a>18, or
a=18 is the maximum value of a for the inequalities to be valid.
We established that to satisfy the quadratic inequality
x^2−11x−210 < 0
-10<x<21
Now substitute this interval into each of the two linear inequalities:
x-8<a and 8-x <a
(-10)-8<a => a>-18
8-(-10)<a => a> 18
(21)-8<a => a>13
8-(21)<a => a>-13
To satisfy all four inequalities, we need a>18.
Finally check at least a point between -10 and 21, say 0.
0-8<18, 8-0<18
so a>18, or
a=18 is the maximum value of a for the inequalities to be valid.
Thanks MathMate :)