# Find the set of all real values of x for which $\dfrac{{8{x^2} + 16x - 51}}{{(2x - 3)(x + 4)}} < 3$.

A) $( - 3,\dfrac{5}{2})$

B) $( - 4, - 3)$

C) Both A and B

D) None of these

Last updated date: 30th Mar 2023

•

Total views: 308.7k

•

Views today: 5.86k

Answer

Verified

308.7k+ views

Hint: we will simplify the inequality and then find the values of the ‘x’ satisfying the inequality by plotting the equation.

Given,

$\dfrac{{8{x^2} + 16x - 51}}{{(2x - 3)(x + 4)}} < 3 \to (1)$

Now, let us simplify the inequality by adding -3 on both sides of equation (1), we get

$

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

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

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

$

Now, solving the numerator, we get

$

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

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

\Rightarrow \dfrac{{2{x^2} + x - 15}}{{(2x - 3)(x + 4)}} < 0 \to (2) \\

$

Now, multiply and divide the L.H.S of equation (2) with $'(2x - 3)(x + 4)'$ , we get

$

\Rightarrow \dfrac{{2{x^2} + x - 15}}{{(2x - 3)(x + 4)}}*(\dfrac{{(2x - 3)(x + 4)}}{{(2x - 3)(x + 4)}}) < 0 \\

\Rightarrow \dfrac{{(2{x^2} + x - 15)((2x - 3)(x + 4))}}{{{{((2x - 3)(x + 4))}^2}}} < 0 \to (3) \\

$

Now, factoring the quadratic equation $2{x^2} + x - 15$ we get $(2x - 5)(x + 3)$ .Therefore, rewriting the equation (3), we get

$ \Rightarrow \dfrac{{(2x - 5)(x + 3)((2x - 3)(x + 4))}}{{{{((2x - 3)(x + 4))}^2}}} < 0 \to (4)$

Now, we can say the denominator is positive, since the square is always positive.

Now, to find the values of x we will only numerator i.e..,

$(2x - 5)(x + 3)((2x - 3)(x + 4)) < 0$

Solving the inequality by plotting, we get

Hence, when $x \in ( - 4, - 3) \cup (\dfrac{3}{2},\dfrac{5}{2})$ , equation (4) will be negative.

So, from the options given, only ‘B’ is suitable. Hence, option ‘B’ i.e.., $( - 4, - 3)$ is the required solution.

Note: Since, the denominator value is positive, therefore we have considered the numerator as negative to find the values of ‘x’ satisfying the given inequality.

Given,

$\dfrac{{8{x^2} + 16x - 51}}{{(2x - 3)(x + 4)}} < 3 \to (1)$

Now, let us simplify the inequality by adding -3 on both sides of equation (1), we get

$

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

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

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

$

Now, solving the numerator, we get

$

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

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

\Rightarrow \dfrac{{2{x^2} + x - 15}}{{(2x - 3)(x + 4)}} < 0 \to (2) \\

$

Now, multiply and divide the L.H.S of equation (2) with $'(2x - 3)(x + 4)'$ , we get

$

\Rightarrow \dfrac{{2{x^2} + x - 15}}{{(2x - 3)(x + 4)}}*(\dfrac{{(2x - 3)(x + 4)}}{{(2x - 3)(x + 4)}}) < 0 \\

\Rightarrow \dfrac{{(2{x^2} + x - 15)((2x - 3)(x + 4))}}{{{{((2x - 3)(x + 4))}^2}}} < 0 \to (3) \\

$

Now, factoring the quadratic equation $2{x^2} + x - 15$ we get $(2x - 5)(x + 3)$ .Therefore, rewriting the equation (3), we get

$ \Rightarrow \dfrac{{(2x - 5)(x + 3)((2x - 3)(x + 4))}}{{{{((2x - 3)(x + 4))}^2}}} < 0 \to (4)$

Now, we can say the denominator is positive, since the square is always positive.

Now, to find the values of x we will only numerator i.e..,

$(2x - 5)(x + 3)((2x - 3)(x + 4)) < 0$

Solving the inequality by plotting, we get

Hence, when $x \in ( - 4, - 3) \cup (\dfrac{3}{2},\dfrac{5}{2})$ , equation (4) will be negative.

So, from the options given, only ‘B’ is suitable. Hence, option ‘B’ i.e.., $( - 4, - 3)$ is the required solution.

Note: Since, the denominator value is positive, therefore we have considered the numerator as negative to find the values of ‘x’ satisfying the given inequality.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

The coordinates of the points A and B are a0 and a0 class 11 maths JEE_Main

Trending doubts

Write an application to the principal requesting five class 10 english CBSE

Tropic of Cancer passes through how many states? Name them.

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE

What is per capita income

Change the following sentences into negative and interrogative class 10 english CBSE

A Short Paragraph on our Country India