What is 2x^2-8<x+4<= 3x-1 ?

The solution to 2x^2-8<x+4<= 3x-1 is 5/2 <= x<(sqrt(97)+1)/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2x^2-8<x+4<= 3x-1

2 x^{2} - 8 < x + 4 \leq 3 x - 1

\frac{5}{2} \leq x < \frac{97 + 1}{4}

Interval Notation

[\frac{5}{2}, \frac{97 + 1}{4})

Decimal Notation

2.5 \leq x < 2.71221 \dots

Solution steps

2 x^{2} - 8 < x + 4 \leq 3 x - 1

a < u \leq b

then

a < u and u \leq b

2 x^{2} - 8 < x + 4 and x + 4 \leq 3 x - 1

2 x^{2} - 8 < x + 4 : \frac{- 97 + 1}{4} < x < \frac{97 + 1}{4}

x + 4 \leq 3 x - 1 : x \geq \frac{5}{2}

Combine the intervals

\frac{- 97 + 1}{4} < x < \frac{97 + 1}{4} and x \geq \frac{5}{2}

Merge Overlapping Intervals

\frac{5}{2} \leq x < \frac{97 + 1}{4}

6 x^{2} - 5 x - 4 = 0

\frac{2}{5} \cdot \frac{2}{3}

2 p^{2} + 7 p - 30 = 0

\frac{3}{4} x - 2 \geq \frac{5}{3} x + 8

x^{2} + 169 = 0

What is 2x^2-8<x+4<= 3x-1 ?
The solution to 2x^2-8<x+4<= 3x-1 is 5/2 <= x<(sqrt(97)+1)/4