What is b^2+9b+7<= 10000 ?

The solution to b^2+9b+7<= 10000 is (-13sqrt(237)-9)/2 <= b<= (13sqrt(237)-9)/2

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b^2+9b+7<= 10000

b^{2} + 9 b + 7 \leq 10000

\frac{- 13 237 - 9}{2} \leq b \leq \frac{13 237 - 9}{2}

Interval Notation

[\frac{- 13 237 - 9}{2}, \frac{13 237 - 9}{2}]

Decimal Notation

- 104.56622 \dots \leq b \leq 95.56622 \dots

Solution steps

b^{2} + 9 b + 7 \leq 10000

Rewrite in standard form

b^{2} + 9 b - 9993 \leq 0

Complete the square

b^{2} + 9 b - 9993 : (b + \frac{9}{2})^{2} - \frac{40053}{4}

(b + \frac{9}{2})^{2} - \frac{40053}{4} \leq 0

Move

\frac{40053}{4}

to the right side

(b + \frac{9}{2})^{2} \leq \frac{40053}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{40053}{4} \leq b + \frac{9}{2} \leq \frac{40053}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{40053}{4} \leq b + \frac{9}{2} and b + \frac{9}{2} \leq \frac{40053}{4}

- \frac{40053}{4} \leq b + \frac{9}{2} : b \geq \frac{- 13 237 - 9}{2}

b + \frac{9}{2} \leq \frac{40053}{4} : b \leq \frac{13 237 - 9}{2}

Combine the intervals

b \geq \frac{- 13 237 - 9}{2} and b \leq \frac{13 237 - 9}{2}

Merge Overlapping Intervals

\frac{- 13 237 - 9}{2} \leq b \leq \frac{13 237 - 9}{2}

ln (2) - ln (4)

3 c + 1 \geq 10

2 (0) + 3

4 n^{2} + 36 n + 81 = 0

\frac{2 \cdot 5 - 4 \cdot 5 ^{2}}{5 ^{2} - 20}

What is b^2+9b+7<= 10000 ?
The solution to b^2+9b+7<= 10000 is (-13sqrt(237)-9)/2 <= b<= (13sqrt(237)-9)/2