|4x+8|<= 9

Solution

∣ 4 x + 8 ∣ \leq 9

- \frac{17}{4} \leq x \leq \frac{1}{4}

Interval Notation

[- \frac{17}{4}, \frac{1}{4}]

Decimal Notation

- 4.25 \leq x \leq 0.25

Solution steps

∣ 4 x + 8 ∣ \leq 9

Apply absolute rule: If

∣ u ∣ \leq a, a > 0

then

- a \leq u \leq a

- 9 \leq 4 x + 8 \leq 9

4 x + 8 \geq - 9 and 4 x + 8 \leq 9

x \geq - \frac{17}{4} and x \leq \frac{1}{4}

Merge Overlapping Intervals

- \frac{17}{4} \leq x \leq \frac{1}{4}

- x^{2} + 5 x < 0

(x + 15)^{4}

\frac{7 !}{3 ! ( 7 - 3 ) !}

10 j - 30 j^{2} + 15 j^{4} - 5 j^{3}

x^{2} - 10 x + 34 = 0