Question

for which value of b the inequality b2 +8b>=9b+14 is correct?

Answer 1

Answer:

b≥ 14 the given inequality is correct

Step-by-step explanation:

Given is

b² + 8b ≥ 9b + 14

Now

we have to find the value of b for this the given inequality above is

b² + 8b ≥ 9b + 14

Now subtracting 9b from both sides of the inequality it becomes

b² + 8b - 9b ≥ 9b - 9b + 14

b²  - b ≥ 14

Now from left side taking b common it becomes

b ( b-1) ≥ 14

Now we have two situations i.e.

either

b ≥ 14      or  b-1 ≥ 14

                    b ≥ 15

As 15 is greater then 14

so both will give the same results

Check:

b² + 8b ≥ 9b + 14

but b=14

(14)²+8(14) ≥ 9(14) + 14

(14)²+8(14) ≥ 9(14) + 14

304 ≥ 140

Answer 2

Solution

${b}^{2} + 8b \geqslant 9b + 14 \\ \\ {b}^{2} + 8b - 9b \geqslant 9b - 9b + 14 \\ \\ {b}^{2} - b \geqslant 14 \\ \\ {b}^{2} - b - 14 \geqslant 0 \\ \\ b_{1,2} = \frac{1 ± \sqrt{1 + 56} }{2} \\ \\ b_{1,2} = \frac{1 ± \sqrt{57} }{2} \\ \\ so \\ \\ b_{1} \geqslant \frac{1 + \sqrt{57} }{2} \\ \\ b_{2} \leqslant \frac{1 - \sqrt{57} }{2}$
The inequality can be hold true for both
b ≥ 4.27
b ≤ 3.27

the inequality can be correct for these values
Hope it helps you.