Question

If one zero of the polynomial \[\left( {{a}^{2}}+9 \right){{x}^{2}}+13x+6a\] is the reciprocal of the other, find the value of a.

Answer 1

Hint: To solve the given question, we will first find out what the zeroes of the polynomial are. Then, we will assume that the zeroes of the given polynomial are \[\alpha \] and \[\beta .\] Then we will develop the relation between \[\alpha \] and \[\beta \] using the information given in the question that \[\alpha \] and \[\beta \] are reciprocal to each other. Then we will make use of the fact that if \[\alpha \] and \[\beta \] are the roots of this equation, then the product of the roots will be equal to the ratio of the coefficient of the constant to the ratio of the coefficient of \[{{x}^{2}}.\] From here, we will get a quadratic equation which we will solve with the help of the factorization method.

Complete step-by-step answer:
Before we solve the given question, we will first find out what zeroes of the polynomial are. The zeroes of the polynomial are those values of the variable which when put in polynomial, it becomes zero. The polynomial given in the question is,
\[P\left( x \right)=\left( {{a}^{2}}+9 \right){{x}^{2}}+13x+6a\]
Now, let us assume that \[\alpha \] and \[\beta \] are the roots of P(x). Now, it is given in the question that \[\alpha \] and \[\beta \] are reciprocal of each other. Thus, we can say that,
\[\beta =\dfrac{1}{\alpha }.....\left( i \right)\]
Now, we know that if \[\alpha \] and \[\beta \] are the zeroes of P(x) then the product of \[\alpha \] and \[\beta \] will be equal to the ratio of the constant term to the coefficient of \[{{x}^{2}}.\] In our case, the constant term is 6a and coefficient of \[{{x}^{2}}\] is \[\left( {{a}^{2}}+9 \right).\] Thus, we will get,
\[\alpha \beta =\dfrac{6a}{{{a}^{2}}+9}.....\left( ii \right)\]
Now, we will substitute the value of \[\beta \] from (i) to (ii). Thus, we will get,
\[\Rightarrow \alpha \times \dfrac{1}{\alpha }=\dfrac{6a}{{{a}^{2}}+9}\]
\[\Rightarrow \dfrac{6a}{{{a}^{2}}+9}=1\]
\[\Rightarrow {{a}^{2}}+9=6a\]
\[\Rightarrow {{a}^{2}}-6a+9=0\]
Now, the above equation is quadratic in a, so we will solve it by factorization method. Thus, we have,
\[\Rightarrow {{a}^{2}}-3a-3a+9=0\]
\[\Rightarrow a\left( a-3 \right)-3\left( a-3 \right)=0\]
\[\Rightarrow \left( a-3 \right)\left( a-3 \right)=0\]
\[\Rightarrow {{\left( a-3 \right)}^{2}}=0\]
\[\Rightarrow a=3\]
Thus the value of a is 3.

Note: The alternate way to solve the question is given below. We know that if there is a quadratic equation of the form \[p{{x}^{2}}+qx+r=0,\] then its roots are given by,
\[x=\dfrac{-q\pm \sqrt{{{q}^{2}}-4pr}}{2p}\]
In our case, q = 13, \[p={{a}^{2}}+9\] and r = 6a. Thus, we have,
\[x=\dfrac{-13\pm \sqrt{{{\left( 13 \right)}^{2}}-4\left( 6a \right)\left( {{a}^{2}}+9 \right)}}{2\left( {{a}^{2}}+9 \right)}\]
Therefore,
\[{{x}_{1}}=\dfrac{-13+\sqrt{169-24a\left( {{a}^{2}}+9 \right)}}{2\left( {{a}^{2}}+9 \right)}\]
\[{{x}_{2}}=\dfrac{-13-\sqrt{169-24a\left( {{a}^{2}}+9 \right)}}{2\left( {{a}^{2}}+9 \right)}\]
Now, \[{{x}_{1}}\] and \[{{x}_{2}}\] are reciprocal of each other, so we have,
\[{{x}_{1}}=\dfrac{1}{{{x}_{2}}}\]
\[\Rightarrow {{x}_{1}}{{x}_{2}}=1\]
So, we have,
\[\left( \dfrac{-13+\sqrt{169-24a\left( {{a}^{2}}+9 \right)}}{2\left( {{a}^{2}}+9 \right)} \right)\left( \dfrac{-13-\sqrt{169-24a\left( {{a}^{2}}+9 \right)}}{2\left( {{a}^{2}}+9 \right)} \right)=1\]
With the help of the identity, \[\left( u+v \right)\left( u-v \right)={{u}^{2}}-{{v}^{2}},\] we will get,
\[\dfrac{{{\left( -13 \right)}^{2}}-{{\left( \sqrt{169-24a\left( {{a}^{2}}+9 \right)} \right)}^{2}}}{{{\left( 2 \right)}^{2}}{{\left( {{a}^{2}}+9 \right)}^{2}}}=1\]
\[\Rightarrow 169-\left( 169-24a\left( {{a}^{2}}+9 \right) \right)=4{{\left( {{a}^{2}}+9 \right)}^{2}}\]
\[\Rightarrow 24a\left( {{a}^{2}}+9 \right)=4{{\left( {{a}^{2}}+9 \right)}^{2}}\]
\[\Rightarrow 6a\left( {{a}^{2}}+9 \right)={{\left( {{a}^{2}}+9 \right)}^{2}}\]
\[\Rightarrow 6a={{a}^{2}}+9\]
\[\Rightarrow {{a}^{2}}-6a+9=0\]
\[\Rightarrow {{\left( a-3 \right)}^{2}}=0\]
\[\Rightarrow a=3\]