Question

122
course
ap precalculus
due tue, oct 14, 2025
1
default gallery

1. $sqrt{3x + 4}<2$

text math set matrix
answer

Explanation:

Step1: Consider the domain

For $\sqrt{3x + 4}$ to be a real - valued function, $3x+4\geq0$. Solving this inequality gives $3x\geq - 4$, so $x\geq-\frac{4}{3}$.

Step2: Square both sides of the inequality

Since both sides of $\sqrt{3x + 4}<2$ are non - negative (because the square root is non - negative and 2 is positive), we can square both sides. We get $3x + 4<4$.

Step3: Solve the resulting linear inequality

Subtract 4 from both sides of $3x + 4<4$: $3x<0$. Then divide both sides by 3 to obtain $x < 0$.

Step4: Combine with the domain

Combining $x\geq-\frac{4}{3}$ and $x < 0$, we get $-\frac{4}{3}\leq x<0$.

Answer:

$-\frac{4}{3}\leq x<0$