The solution to x^2+x-6<3 is (-sqrt(37)-1)/2 <x<(sqrt(37)-1)/2

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x^2+x-6<3

Solution

x^{2} + x - 6 < 3

\frac{- 37 - 1}{2} < x < \frac{37 - 1}{2}

Solution steps

x^{2} + x - 6 < 3

Rewrite in standard form

x^{2} + x - 9 < 0

Complete the square

x^{2} + x - 9 : (x + \frac{1}{2})^{2} - \frac{37}{4}

(x + \frac{1}{2})^{2} - \frac{37}{4} < 0

Move

\frac{37}{4}

to the right side

(x + \frac{1}{2})^{2} < \frac{37}{4}

For

u^{n} < a

, if

n

is even then

- \frac{37}{4} < x + \frac{1}{2} < \frac{37}{4}

a < u < b

then

a < u and u < b

- \frac{37}{4} < x + \frac{1}{2} and x + \frac{1}{2} < \frac{37}{4}

- \frac{37}{4} < x + \frac{1}{2} : x > \frac{- 37 - 1}{2}

x + \frac{1}{2} < \frac{37}{4} : x < \frac{37 - 1}{2}

Combine the intervals

x > \frac{- 37 - 1}{2} and x < \frac{37 - 1}{2}

Merge Overlapping Intervals

\frac{- 37 - 1}{2} < x < \frac{37 - 1}{2}

- 35 y^{4} x^{3} - 30 y^{3} + 50 y^{2}

(\frac{16}{3})^{14}

\frac{3}{5} + 4

6 - 2 x > 4

x^{2} - 64 = 0

What is x^2+x-6<3 ?
The solution to x^2+x-6<3 is (-sqrt(37)-1)/2 <x<(sqrt(37)-1)/2