Question

pls some one solve this..
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Answer 1

We're asked to find the solution to the inequality,

$\longrightarrow\sf{\dfrac{8x^2+16x-51}{(2x-3)(x+4)}<3}$

Subtracting 3 from both sides,

$\longrightarrow\sf{\dfrac{8x^2+16x-51}{(2x-3)(x+4)}-3<0}$

$\longrightarrow\sf{\dfrac{8x^2+16x-51-3(2x-3)(x+4)}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{8x^2+16x-51-3(2x^2+5x-12)}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{8x^2+16x-51-6x^2-15x+36}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{2x^2+x-15}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{2x^2+6x-5x-15}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{2x(x+3)-5(x+3)}{(2x-3)(x+4)}<0}$

$\longrightarrow\sf{\dfrac{(x+3)(2x-5)}{(2x-3)(x+4)}<0\quad\quad\dots(1)}$

Now we solve this inequality using wavy curve method.

Let a function $\sf{f(x)=\dfrac{(x+3)(2x-5)}{(2x-3)(x+4)},\quad\!x\notin\left\{\dfrac{3}{2},\ -4\right\}}$ be defined.

First we've to check possible values for x which make each factorised term in f(x) equal to zero.

We get the following.

• $\sf{x+3=0\quad\implies\quad x=-3}$

• $\sf{2x-5=0\quad\implies\quad x=\dfrac{5}{2}}$

• $\sf{2x-3=0\quad\implies\quad x=\dfrac{3}{2}}$

• $\sf{x+4=0\quad\implies\quad x=-4}$

Then we've to arrange these four values as in a number line, i.e., in ascending order.

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We see $\sf{f(x)>0}$ for $\sf{x>\dfrac{5}{2}.}$

$\setlength{\unitlength}{4mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){15}}\multiput(3,-0.04)(3,0){4}{\circle*{0.3}}\multiput(-1.5,-0.08)(16.8,0){2}{\dots}\put(2.2,-1.5){ \sf{-4} }\put(5.1,-1.5){ \sf{-3} }\put(8.7,-1.5){ \sf{\frac{3}{2}} }\put(11.7,-1.5){ \sf{\frac{5}{2}} }\put(13.1,0.5){+}\end{picture}$

Among the four values mentioned,

• if one value is got odd no. of times, then the function will have the sign, in left of that value in the number line, opposite to that in its right.

• if one value is got even no. of times, then the function will have the same sign, in left of that value in the number line, as that in its right.

Here the case $\sf{x=\dfrac{5}{2}}$ is got only one (odd) time. Therefore $\sf{f(x)<0}$ for $\sf{\dfrac{3}{2}<x<\dfrac{5}{2}.}$

$\setlength{\unitlength}{4mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){15}}\multiput(3,-0.04)(3,0){4}{\circle*{0.3}}\multiput(-1.5,-0.08)(16.8,0){2}{\dots}\put(2.2,-1.5){ \sf{-4} }\put(5.1,-1.5){ \sf{-3} }\put(8.7,-1.5){ \sf{\frac{3}{2}} }\put(11.7,-1.5){ \sf{\frac{5}{2}} }\put(13.2,0.5){+}\put(10.1,0.5){ - }\end{picture}$

The case $\sf{x=\dfrac{3}{2}}$ is got only one (odd) time. Therefore $\sf{f(x)>0}$ for $\sf{-3<x<\dfrac{3}{2}.}$

$\setlength{\unitlength}{4mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){15}}\multiput(3,-0.04)(3,0){4}{\circle*{0.3}}\multiput(-1.5,-0.08)(16.8,0){2}{\dots}\put(2.2,-1.5){ \sf{-4} }\put(5.1,-1.5){ \sf{-3} }\put(8.7,-1.5){ \sf{\frac{3}{2}} }\put(11.7,-1.5){ \sf{\frac{5}{2}} }\multiput(13.2,0.5)(-6.1,0){2}{+}\put(10.1,0.5){ - }\end{picture}$

The case $\sf{x=-3}$ is got only one (odd) time. Therefore $\sf{f(x)<0}$ for $\sf{-4<x<-3.}$

$\setlength{\unitlength}{4mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){15}}\multiput(3,-0.04)(3,0){4}{\circle*{0.3}}\multiput(-1.5,-0.08)(16.8,0){2}{\dots}\put(2.2,-1.5){ \sf{-4} }\put(5.1,-1.5){ \sf{-3} }\put(8.7,-1.5){ \sf{\frac{3}{2}} }\put(11.7,-1.5){ \sf{\frac{5}{2}} }\multiput(13.2,0.5)(-6.1,0){2}{+}\multiput(10.1,0.5)(-6.1,0){2}{ - }\end{picture}$

The case $\sf{x=-4}$ is got only one (odd) time. Therefore $\sf{f(x)>0}$ for $\sf{x<-4.}$

$\setlength{\unitlength}{4mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){15}}\multiput(3,-0.04)(3,0){4}{\circle*{0.3}}\multiput(-1.5,-0.08)(16.8,0){2}{\dots}\put(2.2,-1.5){ \sf{-4} }\put(5.1,-1.5){ \sf{-3} }\put(8.7,-1.5){ \sf{\frac{3}{2}} }\put(11.7,-1.5){ \sf{\frac{5}{2}} }\multiput(13.2,0.5)(-6.1,0){3}{+}\multiput(10.1,0.5)(-6.1,0){2}{ - }\end{picture}$

From this we've to find for which values of x the inequality (1) will be true.

Our inequality is $\sf{f(x)<0}$ which is satisfied by the set of values of x,

$\longrightarrow\sf{\underline{\underline{x\in\big(\!\!-4,\ -3\big)\cup\left(\dfrac{3}{2},\ \dfrac{5}{2}\right)}}}$

This is the answer!