Find the maximum value of a such that the following statement is true:
If ∣x−8∣<a, then x^2−11x−210<0.
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?
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.
Thanks MathMate :)
To find the maximum value of "a" that makes the given statement true, we need to consider the conditions for which the inequality x^2 - 11x - 210 < 0 is satisfied, given that |x - 8| < a.
Let's break this down step by step:
1. Factor the quadratic equation: x^2 - 11x - 210 = (x - 21)(x + 10).
By factoring, we get the roots of the equation, x = 21 and x = -10.
2. Plot the two roots on the number line:
-10 21
We now have two critical points (-10 and 21) on the number line.
3. Find the intervals on the number line:
We need to identify the regions where x^2 - 11x - 210 is negative so that the inequality x^2 - 11x - 210 < 0 holds true.
- Start by examining the interval to the left of -10.
Choosing a test point below -10, let's use x = -20. Replacing x in the equation, we get:
(-20 - 21)(-20 + 10) = (-41)(-10) = 410
Since 410 is positive, we can conclude that x^2 - 11x - 210 is positive for x < -10.
- Next, look at the interval between -10 and 21.
Choosing a test point within the interval, we'll use x = 0. Replacing x in the equation, we get:
(0 - 21)(0 + 10) = (-21)(10) = -210
As -210 is negative, we can conclude that x^2 - 11x - 210 is negative for -10 < x < 21.
- Lastly, we examine the interval to the right of 21.
Choosing a test point above 21, let's use x = 30. Replacing x in the equation, we get:
(30 - 21)(30 + 10) = (9)(40) = 360
Again, we find 360 to be positive, meaning that x^2 - 11x - 210 is positive for x > 21.
4. Combine the intervals:
From step 3, we found that x^2 - 11x - 210 is negative for -10 < x < 21.
Given that |x - 8| < a, we can analyze the intervals for this condition.
Since x = 8 lies within the interval -10 < x < 21, we can see that |x - 8| < a will be satisfied when -10 < x < 21.
To maintain the inequality |x - 8| < a, we need to find the maximum value of "a" that satisfies the interval.
The maximum value of a can then be calculated as the minimum difference between 8 and the endpoints of the interval:
a = min(8 - (-10), 21 - 8)
= min(18, 13)
= 13
Hence, the maximum value of a is 13, which ensures that if |x - 8| < 13, then x^2 - 11x - 210 < 0.