Question

Solve the following inequality : \[({x^2} - x - 1) \times ({x^2} - x - 7) < - 5\]

Answer 1

Hint: We have to find the value of x from the given expression of inequality \[({x^2} - x - 1) \times ({x^2} - x - 7) < - 5\]. We solve this question using the concept of solving linear equations of inequality. We will first simplify the terms of the inequality by multiplying the terms of the expression. Then we would factorise the expression into its factors. And then making the intervals of the values of \[x\], and on further solving the expression of the inequality we will get the range for the value of \[x\] for which it satisfies the given expression .

Complete step-by-step solution:
Given :
\[({x^2} - x - 1) \times ({x^2} - x - 7) < - 5\]
Simplifying the inequality by multiplication of the expression , we can write the expression of the inequality as :
\[{x^4} - {x^3} - {x^2} - {x^3} + {x^2} + x - 7{x^2} + 7x + 7 < - 5\]
\[{x^4} - 2{x^3} - 7{x^2} + 8x + 12 < 0\]
Now , by hit and trial method , we can find the factors of the expression of inequality as :
For finding the factors , we can write the equation of inequality as :
\[{x^4} - 2{x^3} - 7{x^2} + 8x + 12 = 0\]
Putting \[x{\text{ }} = - 1\] in the expression , we get the value as :
\[{x^4} - 2{x^3} - 7{x^2} + 8x + 12 = {( - 1)^4} - 2{( - 1)^3} - 7{( - 1)^2} + 8( - 1) + 12\]
\[{x^4} - 2{x^3} - 7{x^2} + 8x + 12 = 1 + 2 - 7 - 8 + 12\]
On solving , we get
\[{x^4} - 2{x^3} - 7{x^2} + 8x + 12 = 0\]
Hence , \[x{\text{ }} = - 1\] is a factor of the given expression .
Now , we can split the expression as :
\[{x^4} - 3{x^3} + {x^3} - 4{x^2} - 3{x^2} + 12x - 4x + 12 = 0\]
Taking the terms common such that , we get \[\left( {x + 1} \right)\] common
\[{x^3}(x + 1) - 3{x^2}(x + 1) - 4x(x + 1) + 12(x + 1) = 0\]
Taking \[\left( {x + 1} \right)\] common , we get
\[({x^3} - 3{x^2} - 4x + 12) \times (x + 1) = 0\]
Now , we have to factor the expression \[({x^3} - 3{x^2} - 4x + 12) = 0\] .
Putting \[x = 2\] in the expression by hit and trial method , we get the value as :
\[({x^3} - 3{x^2} - 4x + 12) = {2^3} - 3{(2)^2} - 4(2) + 12\]
\[({x^3} - 3{x^2} - 4x + 12) = 8 - 12 - 8 + 12\]
\[({x^3} - 3{x^2} - 4x + 12) = 0\]
Hence , \[x - 2\] is a factor of the expression .
Now , we can split the expression as :
\[{x^3} - 2{x^2} - {x^2} - 6x + 2x + 12 = 0\]
Taking the terms common such that , we get \[\left( {x - 2} \right)\] common
\[{x^2}(x - 2) - x(x - 2) - 6(x - 2) = 0\]
Taking \[\left( {x - 2} \right)\] common , we get
\[({x^2} - x - 6) \times (x - 2) = 0\]
Now , we have to factor the expression \[({x^2} - x - 6) = 0\] .
Now splitting the equation , we can write the expression as :
\[{x^2} - 3x + 2x - 6 = 0\]
\[x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\]
Also , we can write it as :
\[\left( {x + 2} \right)\left( {x - 3} \right) = 0\]
So , the other two factors are \[x = 3\] and \[x = - 2\] .
Now , the factors of the inequality are \[ - 2\] , \[ - 1\] , \[2\] and \[3\] .
We can write the expression of inequality as :
\[\left( {x + 1} \right)\left( {x - 2} \right)\left( {x + 2} \right)\left( {x - 3} \right) < 0\]
Now , we can write the intervals as :
\[\left( { - \infty , - 2} \right),\left( { - 2, - 1} \right),\left( { - 1,2} \right),\left( {2,3} \right),\left( {3,\infty } \right)\]
Now for the range we will check the inequality for a value between each of the intervals .
Put value \[\left( { - \infty , - 2} \right)\] , \[x = - 3\] in the inequality , we get the value of inequality as :
\[\left( { - 3 + 1} \right)\left( { - 3 - 2} \right)\left( { - 3 + 2} \right)\left( { - 3 - 3} \right)\]
\[\left( { - 2} \right)\left( { - 5} \right)\left( { - 1} \right)\left( { - 6} \right) > 0\]
The value obtained would be positive .
As , the value of the inequality should be negative so we can’t have a value of \[x \in \left( { - \infty , - 2} \right)\] .
Put value \[\left( { - 2, - 1} \right)\] , \[x = \dfrac{{ - 3}}{2}\] in the inequality , we get the value of inequality as :
\[\left( {\dfrac{{ - 3}}{2} + 1} \right)\left( {\dfrac{{ - 3}}{2} - 2} \right)\left( {\dfrac{{ - 3}}{2} + 2} \right)\left( {\dfrac{{ - 3}}{2} - 3} \right)\]
\[\left( {\dfrac{{ - 1}}{2}} \right)\left( {\dfrac{{ - 7}}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( {\dfrac{{ - 9}}{2}} \right) < 0\]
The value obtained would be negative.
As , the value of the inequality should be negative so we can have a value of \[x \in \left( { - 2, - 1} \right)\] .
Put value \[\left( { - 1,2} \right)\] , \[x = 0\] in the inequality , we get the value of inequality as:
\[\left( {0 + 1} \right)\left( {0 - 2} \right)\left( {0 + 2} \right)\left( {0 - 3} \right)\]
\[\left( 1 \right)\left( { - 2} \right)\left( 2 \right)\left( { - 3} \right) > 0\]
The value obtained would be positive .
As , the value of the inequality should be negative so we can’t have a value of \[x \in \left( { - 1,2} \right)\] .
Put value \[\left( {2,3} \right)\] , \[x = \dfrac{5}{2}\] in the inequality , we get the value of inequality as :
\[\left( {\dfrac{5}{2} + 1} \right)\left( {\dfrac{5}{2} - 2} \right)\left( {\dfrac{5}{2} + 2} \right)\left( {\dfrac{5}{2} - 3} \right)\]
\[\left( {\dfrac{7}{2}} \right)\left( {\dfrac{1}{2}} \right)\left( {\dfrac{9}{2}} \right)\left( {\dfrac{{ - 1}}{2}} \right) < 0\]
The value obtained would be negative.
As , the value of the inequality should be negative so we can have a value of \[x \in \left( {2,3} \right)\] .
Put value \[\left( {3,\infty } \right)\] , \[x = 4\] in the inequality , we get the value of inequality as :
\[\left( {4 + 1} \right)\left( {4 - 2} \right)\left( {4 + 2} \right)\left( {4 - 3} \right)\]
\[\left( 5 \right)\left( 2 \right)\left( 6 \right)\left( 1 \right) > 0\]
The value obtained would be positive .
As , the value of the inequality should be negative so we can’t have a value of \[x \in \left( {3,\infty } \right)\] .
Hence , the solution of \[x\] from the inequality \[({x^2} - x - 1) \times ({x^2} - x - 7) < - 5\] is \[\left( { - 2, - 1} \right) \cup \left( {2,3} \right)\].

Note: We must take care about the sign and symbols of the inequality , as a slight change causes major errors in the solution . The solution of the range of the inequality states that each and every value which lies in that particular range satisfies the given equation . The round bracket \[\left( {} \right)\] in the value of the range states that the end elements I.e. \[ - 2\] , \[2\] , \[3\] and \[ - 1\] in this question will not satisfy the given expression whereas the square bracket \[\left[ {} \right]\] states that the end elements of the range will satisfy the given expression.