Question

Solve the equation \[6{x^4} - 13{x^3} - 35{x^2} - x + 3 = 0\], having given that one roots is \[2 - \sqrt 3 \]

Answer 1

Hint: Here we need to solve the given polynomial equation. We will find the one more root of the polynomial using this concept that the complex roots are present in conjugate pairs. We will write the given equation as the product of two quadratic equations. So, we will find the first quadratic equation using these two roots. We will assume the second quadratic equation as we will write the given polynomial as the product of two quadratic equations. Then we will compare the coefficients and from there, we will get the second quadratic equation and hence, the two more roots will be obtained from there.

Complete step-by-step answer:
Here we need to solve the given polynomial equation.
One complex root of the polynomial is given i.e. \[2 - \sqrt 3 \].
We know that complex roots always present in conjugate pairs. So another root of the polynomial will be equal to \[2 + \sqrt 3 \].
As the degree of the given polynomial is 4, so we can write it as the product of two quadratic equations.
We will find the first quadratic equation using these two roots.
We know the general form of the quadratic equation is given by
\[{x^2} - \] (sum of products)\[x + \] (product of roots) \[ = 0\] ………. \[\left( 1 \right)\]
So we will first find the sum of these two complex roots.
Sum of the two roots \[ = 2 + \sqrt 3 + 2 - \sqrt 3 = 4\]
Now, we will find the product of these two complex roots.
Product of the two roots \[ = \left( {2 + \sqrt 3 } \right) \times \left( {2 - \sqrt 3 } \right) = 4 - 3 = 1\]
Now, we will substitute the value of the sum of roots and the product of roots in equation \[\left( 1 \right)\]. Therefore, we get
\[{x^2} - {\rm{4}}x + {\rm{1}} = 0\] ……….. \[\left( 2 \right)\]
Let the second quadratic equation be \[a{x^2} + bx + c\].
Now, we will write the given polynomial as the product of two quadratic equations.
\[6{x^4} - 13{x^3} - 35{x^2} - x + 3 = \left( {{x^2} - {\rm{4}}x + {\rm{1}}} \right)\left( {a{x^2} + bx + c} \right)\]
Now, we will multiply these two equations.
\[ \Rightarrow 6{x^4} - 13{x^3} - 35{x^2} - x + 3 = a{x^4} - 4a{x^3} + a{x^2} + b{x^3} - 4b{x^2} + bx + c{x^2} - 4cx + c\]
Now, we will add the like terms here.
\[ \Rightarrow 6{x^4} - 13{x^3} - 35{x^2} - x + 3 = a{x^4} + \left( { - 4a + b} \right){x^3} + \left( {a - 4b + c} \right){x^2} + \left( {b - 4c} \right)x + c\]
Now, we will compare the coefficients of the polynomial of both sides.
\[a = 6\]
\[ - 4a + b = - 13\] ………… \[\left( 3 \right)\]
\[a - 4b + c = - 13\] …………. \[\left( 4 \right)\]
\[b - 4c = - 1\] …….. \[\left( 5 \right)\]
\[c = 3\]
Now, we will substitute the value of \[\left( 3 \right)\] in equation 3.
\[ - 4 \times 6 + b = - 13\]
On multiplying the terms, we get
\[ \Rightarrow - 24 + b = - 13\]
Now, we will add 24 to both sides of the equation.
\[\begin{array}{l} \Rightarrow - 24 + b + 24 = - 13 + 24\\ \Rightarrow b = 11\end{array}\]
Therefore, the second quadratic equation becomes:
\[6{x^2} + 11x + 3 = 0\]
If we find the roots of this quadratic equation, we will get all the roots of the polynomial.
Now, we will find the roots of this quadratic equation by factoring it.
First, we will split the middle term of the quadratic equation \[6{x^2} + 11x + 3\] into two new terms. We need to split the terms such that the coefficients of the two terms are the factors of the term obtained from the multiplication of the coefficient of \[x{}^2\] and the constant term.
Multiplication of the coefficient of \[x{}^2\] and the constant term \[ = 18\]
The factors of 18 are 9 and 2. Therefore, we will split the middle term as;
\[6{x^2} + 9x + 2x + 3 = 0\]
Now, we will factorize the first two terms and last two terms separately.
\[ \Rightarrow 3x\left( {2x + 3} \right) + \left( {2x + 3} \right) = 0\]
We can see that both new terms formed have a common factor.
Therefore,
\[ \Rightarrow \left( {3x + 1} \right)\left( {2x + 3} \right) = 0\]
This is possible when \[3x + 1 = 0\] and \[2x + 3 = 0\]
On further simplification, we get
\[\begin{array}{l} \Rightarrow 3x = - 1\\ \Rightarrow x = \dfrac{{ - 1}}{3}\end{array}\]
And also
\[\begin{array}{l} \Rightarrow 2x = - 3\\ \Rightarrow x = \dfrac{{ - 3}}{2}\end{array}\]
Now, we have got all the roots of the polynomial.
Hence, the required roots of the polynomial are \[2 + \sqrt 3 \] , \[2 - \sqrt 3 \], \[\dfrac{{ - 1}}{3}\] and \[\dfrac{{ - 3}}{2}\]
Hence, this is the required solution of this polynomial.

Note: An equation having its degree 2 is called Quadratic Equation. A quadratic equation has two solutions as the highest degree of variable is 2.
For Example - \[{x^2} + x + 1 = 0\] ,
\[2{y^2} + 3y - 5 = 0\] etc.
The General Form of Quadratic Equation is \[a{x^2} + b{y^2} + 2hxy + 2gx + 2fy + c = 0\].
This equation also represents a Circle, Pair of Straight Lines, Ellipse, Parabola and Hyperbola.