What is 2x^2+5x<-x^2+1 ?

The solution to 2x^2+5x<-x^2+1 is (-sqrt(37)-5)/6 <x<(sqrt(37)-5)/6

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2x^2+5x<-x^2+1

2 x^{2} + 5 x < - x^{2} + 1

\frac{- 37 - 5}{6} < x < \frac{37 - 5}{6}

Interval Notation

(\frac{- 37 - 5}{6}, \frac{37 - 5}{6})

Decimal Notation

- 1.84712 \dots < x < 0.18046 \dots

Solution steps

2 x^{2} + 5 x < - x^{2} + 1

Rewrite in standard form

3 x^{2} + 5 x - 1 < 0

Complete the square

3 x^{2} + 5 x - 1 : 3 (x + \frac{5}{6})^{2} - \frac{37}{12}

3 (x + \frac{5}{6})^{2} - \frac{37}{12} < 0

Move

\frac{37}{12}

to the right side

3 (x + \frac{5}{6})^{2} < \frac{37}{12}

Divide both sides by

3

(x + \frac{5}{6})^{2} < \frac{37}{36}

For

u^{n} < a

, if

n

is even then

- \frac{37}{36} < x + \frac{5}{6} < \frac{37}{36}

a < u < b

then

a < u and u < b

- \frac{37}{36} < x + \frac{5}{6} and x + \frac{5}{6} < \frac{37}{36}

- \frac{37}{36} < x + \frac{5}{6} : x > \frac{- 37 - 5}{6}

x + \frac{5}{6} < \frac{37}{36} : x < \frac{37 - 5}{6}

Combine the intervals

x > \frac{- 37 - 5}{6} and x < \frac{37 - 5}{6}

Merge Overlapping Intervals

\frac{- 37 - 5}{6} < x < \frac{37 - 5}{6}

(x + 2) = 0

\frac{x}{2} + \frac{x}{3} \geq 5

∣ z + 8 ∣ = ∣ z - 5 ∣

m^{2} - 10 m + 16

25 x^{2} - 10 x + 1 = 0

What is 2x^2+5x<-x^2+1 ?
The solution to 2x^2+5x<-x^2+1 is (-sqrt(37)-5)/6 <x<(sqrt(37)-5)/6