Question

If $\left| {2x - 3} \right| < \left| {x + 5} \right|$, then $x$ belongs to :
A. $\left( { - 3,5} \right)$
B. $\left( {5,9} \right)$
C. $\left( {\dfrac{{ - 2}}{3},8} \right)$
D. $\left( { - 8,\dfrac{2}{3}} \right)$

Answer 1

Hint: First of all this is a very simple and a very easy problem. This problem deals with basic concepts in algebraic mathematics. In order to solve this problem we should be able to write mathematical expressions from the given data, and also we should be able to equate these mathematical expressions to the given relatable data, and obtain correct results. We should be familiar in solving quadratic expressions as well.

Complete step-by-step solution:
Given an inequation above, which is given by $\left| {2x - 3} \right| < \left| {x + 5} \right|$,
We have to find the interval of $x$, from the above inequation.
This can be in 2 ways, either solving by removing the modulus of the both the expressions in the above inequation.
Another way to do this that by squaring on both sides of the inequation, as given below:
$ \Rightarrow \left| {2x - 3} \right| < \left| {x + 5} \right|$
Squaring the inequation on both sides as given below:
\[ \Rightarrow {\left| {2x - 3} \right|^2} < {\left| {x + 5} \right|^2}\]
\[ \Rightarrow 4{x^2} + 9 - 2(2x)(3) < {x^2} + 25 + 2(5)(x)\]
Simplifying the expressions in the above equations, as given below:
\[ \Rightarrow 4{x^2} + 9 - 12x < {x^2} + 25 + 10x\]
\[ \Rightarrow 3{x^2} - 22x - 16 < 0\]
Hence by rearranging the terms we obtained a quadratic expression, solving the quadratic expression as given below:
\[ \Rightarrow x < \dfrac{{ - ( - 22) \pm \sqrt {{{(22)}^2} - 4(3)( - 16)} }}{{2(3)}}\]
\[ \Rightarrow x < \dfrac{{22 \pm \sqrt {676} }}{6}\]
On further simplification, as given below:
\[ \Rightarrow x < \dfrac{{22 \pm 26}}{6}\]
\[ \Rightarrow x < \dfrac{{22 + 26}}{6};x < \dfrac{{22 - 26}}{6}\]
\[ \Rightarrow x < 8;x < \dfrac{{ - 2}}{3}\]
Hence the roots can be written as given below:
\[ \Rightarrow (x - 8)(x + \dfrac{2}{3}) < 0\]
$\therefore x \in \left( {\dfrac{{ - 2}}{3},8} \right)$

The interval of $x$ belongs to $\left( {\dfrac{{ - 2}}{3},8} \right)$

Note: While solving such this problem we should understand that, there is a quadratic expression, and to solve any quadratic expression which is in the form of $a{x^2} + bx + c = 0$, the roots of the quadratic expression are given by the formula $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where here $a$ is the coefficient of the term ${x^2}$, and $b$ is the coefficient of the term $x$, and $c$ is the constant in the quadratic expression.