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Question: If \(\alpha \) is a real root of \(2{x^3} - 3{x^2} + 6x + 6 = 0\) , then find \(\left[ \alpha \right...

If α\alpha is a real root of 2x3−3x2+6x+6=02{x^3} - 3{x^2} + 6x + 6 = 0 , then find [α]\left[ \alpha \right] where [⋅]\left[ \cdot \right] denotes the greatest integer function.

Explanation

Solution

In the question, we are given with a polynomial equation 2x3−3x2+6x+6=02{x^3} - 3{x^2} + 6x + 6 = 0 and the root of the polynomial is α\alpha . Now, we will differentiate the given equation. After differentiating, we have to find the discriminant of the function. As the discriminant of f′(x)<0f'\left( x \right) < 0 . Hence, f′(x)>0f'\left( x \right) > 0 . So, the function f(x)f\left( x \right) is an increasing function which means that the function has only one root. Now, we can use hit and trial method to find the real root. When we will have the function value zero, that means that particular value is its root. After that, we will have a range that will contain the root. We can easily find the greatest integer function as it is the nearest integer value to the range.

Complete step-by-step solution:
In the above question, we have to find the root of the polynomial 2x3−3x2+6x+6=02{x^3} - 3{x^2} + 6x + 6 = 0 .
Now, differentiate the above equation:
⇒f′(x)=ddx(2x3−3x2+6x+6) ⇒f′(x)=6x2−6x+6 ⇒f′(x)=6(x2−x+1) \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {2{x^3} - 3{x^2} + 6x + 6} \right) \\\ \Rightarrow f'\left( x \right) = 6{x^2} - 6x + 6 \\\ \Rightarrow f'\left( x \right) = 6\left( {{x^2} - x + 1} \right)

As we can see that the discriminant of f′(x)f'\left( x \right) is less than zero.
So, the differentiation of function which is f′(x)f'\left( x \right)will always be greater than 0.
Thus, f(x)f\left( x \right) is an increasing function and it has only one root.
Now, we will use hit and trial method to find the root of the given equation.
Let x=−1x = - 1
Then, function will be:
⇒f(−1)=2(−1)3−3(−1)2+6(−1)+6 ⇒f(−1)=−2−3−6+6 ⇒f(−1)=−5 \Rightarrow f\left( { - 1} \right) = 2{\left( { - 1} \right)^3} - 3{\left( { - 1} \right)^2} + 6\left( { - 1} \right) + 6 \\\ \Rightarrow f\left( { - 1} \right) = - 2 - 3 - 6 + 6 \\\ \Rightarrow f\left( { - 1} \right) = - 5
Let x=0x = 0
Then, function will be:
⇒f(0)=2(0)3−3(0)2+6(0)+6 ⇒f(0)=6 \Rightarrow f\left( 0 \right) = 2{\left( 0 \right)^3} - 3{\left( 0 \right)^2} + 6\left( 0 \right) + 6 \\\ \Rightarrow f\left( 0 \right) = 6
As we can see from the above obtained values that f(−1)<0f\left( { - 1} \right) < 0 and f(0)>0f\left( 0 \right) > 0 . So, the function value is zero in between -1 and 0.
This implies that α\alpha should lie between -1 and 0.

Hence, the greatest integer function of α\alpha is -1 or [α]=−1\left[ \alpha \right] = - 1.

Note: In these types of questions, when the polynomial and its root is given then our first approach is to find whether the given polynomial has one or more than one root. This can be done by differentiating the polynomial and by finding its discriminant. If the function has only one root, then we can use hit and trial method to find the range of the root.