If sqrt{9 x^2+6 x+1}

Question

(A) Given the inequality: $\sqrt{9 x^2+6 x+1} < (2-x)$ Since $\sqrt{(3x+1)^2} = |3x+1|$,the inequality becomes $|3x+1| < 2-x$. For the square root to

Vedclass · Accepted Answer

$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$

Vedclass · Answer

(A) Given the inequality: $\sqrt{9 x^2+6 x+1} Since $\sqrt{(3x+1)^2} = |3x+1|$,the inequality becomes $|3x+1| For the square root to be defined,we must have $9x^2+6x+1 \ge 0$,which is always true as $(3x+1)^2 \ge 0$. Also,for the inequality to hold,the right side must be positive: $2-x > 0 \implies x Now,solve $|3x+1| This is equivalent to $-(2-x) Case $1$: $3x+1 Case $2$: $3x+1 > -(2-x) \implies 3x+1 > -2+x \implies 2x > -3 \implies x > -\frac{3}{2}$. Combining these,we get $-\frac{3}{2} Thus,$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$.

If $\sqrt{9 x^2+6 x+1} < (2-x)$,then:

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