Question

 If two zeroes of the polynomial \[{{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35\] are \[2\pm \sqrt{3}\], find the other zeroes.

Answer 1

Hint: In this question, we need to consider any of the two formulae relating the roots of the polynomial and the coefficients of the polynomial. Then substitute the given two roots in the following equations which have two unknowns. Now, solving those two equations gives the value of the other two roots.

Complete step by step answer:
BIQUADRATIC POLYNOMIAL: A bi quadratic polynomial is a polynomial having degree four.
The general form of a bi quadratic polynomial is given by \[{{a}_{0}}{{x}^{4}}+{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}_{3}}x+{{a}_{4}}=0\]
ZERO OF A POLYNOMIAL: A real number \[\alpha \] is a zero of the polynomial p(x), if and only if \[p\left( \alpha \right)=0\]
Now, let us assume the zeroes of the bi quadratic polynomial as
\[\alpha ,\beta ,\gamma ,\delta \]
Now, from the bi quadratic polynomial \[{{a}_{0}}{{x}^{4}}+{{a}_{1}}{{x}^{3}}+{{a}_{2}}{{x}^{2}}+{{a}_{3}}x+{{a}_{4}}=0\] we have,
Sum of the roots of the bi quadratic polynomial is given by
\[\alpha +\beta +\gamma +\delta =\dfrac{-{{a}_{1}}}{{{a}_{0}}}\]
Now, sum of the product of the two roots at a time is given by
\[\alpha \beta +\alpha \gamma +\alpha \delta +\beta \gamma +\beta \delta +\gamma \delta =\dfrac{{{a}_{2}}}{{{a}_{0}}}\]
Now, sum of the product of the roots taken three at a time is given by
\[\alpha \beta \gamma +\alpha \beta \delta +\beta \gamma \delta +\alpha \gamma \delta =\dfrac{-{{a}_{3}}}{{{a}_{0}}}\]
Now, product of all the roots is given by
\[\alpha \beta \gamma \delta =\dfrac{{{a}_{4}}}{{{a}_{0}}}\]
Now, let us assume the other two roots of the given bi quadratic equation as a, b
Let us now compare the given bi quadratic polynomial with the general equation then we get,
\[\Rightarrow {{x}^{4}}-6{{x}^{3}}-26{{x}^{2}}+138x-35=0\]
\[{{a}_{0}}=1,{{a}_{1}}=-6,{{a}_{2}}=-26,{{a}_{3}}=138,{{a}_{4}}=-35\]
Now, on comparing the given roots with the roots in the formulae we get,
\[\alpha =2+\sqrt{3},\beta =2-\sqrt{3},\gamma =a,\delta =b\]
Now, from the formula of sum of the roots on substituting the above values we get,
\[\Rightarrow \alpha +\beta +\gamma +\delta =\dfrac{-{{a}_{1}}}{{{a}_{0}}}\]
\[\Rightarrow 2+\sqrt{3}+2-\sqrt{3}+a+b=\dfrac{-\left( -6 \right)}{1}\]
Now, on further simplification we get,
\[\Rightarrow 4+a+b=6\]
Now, on rearranging the terms we get,
\[\Rightarrow a+b=2........\left( 1 \right)\]
Let us now consider the product of the roots formula and substitute the values.
\[\Rightarrow \alpha \beta \gamma \delta =\dfrac{{{a}_{4}}}{{{a}_{0}}}\]
\[\Rightarrow \left( 2+\sqrt{3} \right)\times \left( 2-\sqrt{3} \right)\times a\times b=\dfrac{-35}{1}\]
Now, this can be further written as
\[\Rightarrow \left( 4-3 \right)\times a\times b=-35\]
Now, this can be further simplified and written as
\[\Rightarrow ab=-35\]
Now, let us write b in terms of a
\[\therefore b=\dfrac{-35}{a}\]
Now, substitute this value of b in the equation (1)
\[\Rightarrow a+\dfrac{-35}{a}=2\]
Now, this can be further written as
\[\Rightarrow {{a}^{2}}-35=2a\]
Let us now rearrange the terms and form the quadratic equation.
\[\Rightarrow {{a}^{2}}-2a-35=0\]
Now, the above quadratic equation can be further written as
\[\Rightarrow {{a}^{2}}+5a-7a-35=0\]
Now, the above quadratic equation can be further factored as
\[\Rightarrow a\left( a+5 \right)-7\left( a+5 \right)=0\]
\[\Rightarrow \left( a+5 \right)\left( a-7 \right)=0\]
Now, the possible values of a are
\[\therefore a=-5,7\]
Now, as we know that
\[b=\dfrac{-35}{a}\]
\[\therefore b=7,-5\]
Hence, the other two roots are -5, 7

Note:
Instead of using the sum of the roots and the product of the roots formulae we can use any of the four formulae to get the equations involving a and b. As there are only two unknowns so considering any two equations will be sufficient to get the values of a and b.
It is important to note that the other two roots will be the same for any value of a because if a is -5 then b will be 7 or else if a is 7 then b will be -5 which means the other two roots will be the same at the end.
We can also use the direct formula to get the possible values of a instead of using the factorisation method. Both the methods give the same result.