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Find the least value of $b$ such that $b^2+2b-15 \le 0$.

Guest Mar 19, 2017
 #1
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Find the least value of b such that b2+ 2b - 15 ≤ 0 .

 

First let's find what value of b makes

b2+ 2b - 15 = 0

 

Factor it.

(b - 3)(b + 5) = 0

 

Set each factor equal to zero.

b - 3 = 0          and          b + 5 = 0

b = 3               and          b = -5

 

Now to find out what all values of b makes it less than 0,

let's test a point in the middle of -5 and 3, say 0, and see if that makes the inequality true.

 

02 + 2(0) - 15 ≤ 0

- 15 ≤ 0  true

 

Just to be safe let's test a point less than -5, say -6, and a point greater than 3, say 4.

36 - 12 - 15 ≤ 0

9 ≤ 0  false

16 + 8 - 15 ≤ 0

9 ≤ 0  false

 

So all this means that -5 ≤ b ≤ 3

hectictar  Mar 19, 2017

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