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

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

(A) Given the inequality: $\sqrt{9x^2+6x+1} Since $9x^2+6x+1 = (3x+1)^2$,the expression becomes $\sqrt{(3x+1)^2} This simplifies to $|3x+1| For the square root to be defined,we must have $2-x > 0$,which implies $x Now,solving the absolute value inequality $|3x+1| $-(2-x) Case $1$: $3x+1 $4x Case $2$: $3x+1 > -(2-x)$$3x+1 > -2+x$$2x > -3 \Rightarrow x > -\frac{3}{2}$Combining these,we get $-\frac{3}{2} Thus,$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$.

Class 11 Mathematics Linear Inequalities Word problem of Linear inequalities

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

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A
$x \in \left(-\frac{3}{2}, \frac{1}{4}\right)$
B
$x \in \left(-\frac{3}{2}, \frac{1}{4}\right]$
C
$x \in \left[-\frac{3}{2}, \frac{1}{4}\right)$
D
$x < \frac{1}{4}$

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

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The length of a rectangle is five times the breadth. If the minimum perimeter of the rectangle is $180 \ cm$,then:

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