The solution to 2x^2<9-6x is (-3sqrt(3)-3)/2 <x<(3sqrt(3)-3)/2

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2x^2<9-6x

Solution

2 x^{2} < 9 - 6 x

\frac{- 3 3 - 3}{2} < x < \frac{3 3 - 3}{2}

Interval Notation

(\frac{- 3 3 - 3}{2}, \frac{3 3 - 3}{2})

Decimal Notation

- 4.09807 \dots < x < 1.09807 \dots

Solution steps

2 x^{2} < 9 - 6 x

Rewrite in standard form

2 x^{2} + 6 x - 9 < 0

Complete the square

2 x^{2} + 6 x - 9 : 2 (x + \frac{3}{2})^{2} - \frac{27}{2}

2 (x + \frac{3}{2})^{2} - \frac{27}{2} < 0

Move

\frac{27}{2}

to the right side

2 (x + \frac{3}{2})^{2} < \frac{27}{2}

Divide both sides by

2

(x + \frac{3}{2})^{2} < \frac{27}{4}

For

u^{n} < a

, if

n

is even then

- \frac{27}{4} < x + \frac{3}{2} < \frac{27}{4}

a < u < b

then

a < u and u < b

- \frac{27}{4} < x + \frac{3}{2} and x + \frac{3}{2} < \frac{27}{4}

- \frac{27}{4} < x + \frac{3}{2} : x > \frac{- 3 3 - 3}{2}

x + \frac{3}{2} < \frac{27}{4} : x < \frac{3 3 - 3}{2}

Combine the intervals

x > \frac{- 3 3 - 3}{2} and x < \frac{3 3 - 3}{2}

Merge Overlapping Intervals

\frac{- 3 3 - 3}{2} < x < \frac{3 3 - 3}{2}

5 x^{2} = 50 0^{2}

3 ∣ x - 5 ∣ + 12 \geq 15

x^{2} + 3 x = - 2

- 126 = 14 k

y^{2} - 20 < - 4

What is 2x^2<9-6x ?
The solution to 2x^2<9-6x is (-3sqrt(3)-3)/2 <x<(3sqrt(3)-3)/2