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Question: If one zero of the polynomial \(\left( {{a}^{2}}+2 \right){{x}^{2}}+10x+\left. 3a \right)\) is the r...

If one zero of the polynomial (a2+2)x2+10x+3a)\left( {{a}^{2}}+2 \right){{x}^{2}}+10x+\left. 3a \right) is the reciprocal of the other. Find a

Explanation

Solution

We need to find the value of the variable a if one zero of the polynomial is the reciprocal of the other. We start to solve the given question by considering the zeroes of the polynomial as α\alpha , 1α\dfrac{1}{\alpha } . Then, we use the product of the zeroes formula to get the desired result.

Complete step-by-step solution:
We are given a polynomial and are asked to find the value of the variable a given that one zero of the polynomial is the reciprocal of the other.
We will be solving the given question using the formula of product of the zeroes and the concept of quadratic equations.
The Quadratic equations are the polynomials with degree two. The quadratic equation will always have two roots. The roots may be real or imaginary.
The quadratic equation in standard form is given as follows,
⇒f(x)=px2+qx+r\Rightarrow f\left( x \right)=p{{x}^{2}}+qx+r
Here,
p is the coefficient of x2{{x}^{2}}
q is the coefficient of xx
r is the constant term
According to our question,
The given quadratic equation is (a2+2)x2+10x+3a)\left( {{a}^{2}}+2 \right){{x}^{2}}+10x+\left. 3a \right)
Let the first zero of the polynomial be α\alpha
The reciprocal of the number xx is 1x\dfrac{1}{x} .
Following the same,
The other zero of the polynomial will be 1α\dfrac{1}{\alpha }
Comparing the quadratic equation (a2+2)x2+10x+3a)\left( {{a}^{2}}+2 \right){{x}^{2}}+10x+\left. 3a \right) with the standard form px2+qx+rp{{x}^{2}}+qx+r , we get,
p=a2+2p={{a}^{2}}+2 ;
q=10q=10 ;
r=3ar=3a
From the concept of quadratic equations, the product of the zeroes of a polynomial px2+qx+rp{{x}^{2}}+qx+r is given as follows,
⇒product of zeroes=rp\Rightarrow \text{product of zeroes}=\dfrac{r}{p}
From the above,
we know that the product of zeroes is α×1α\alpha \times \dfrac{1}{\alpha }
Substituting the values in the above equation, we get,
⇒α×1α=3aa2+2\Rightarrow \alpha \times \dfrac{1}{\alpha }=\dfrac{3a}{{{a}^{2}}+2}
Simplifying the above equation, we get,
⇒1=3aa2+2\Rightarrow 1=\dfrac{3a}{{{a}^{2}}+2}
Moving the term a2+2{{a}^{2}}+2 to the other side of the equation, we get,
⇒a2+2=3a\Rightarrow {{a}^{2}}+2=3a
Moving the term 3a3a to the other side of the equation, we get,
⇒a2−3a+2=0\Rightarrow {{a}^{2}}-3a+2=0
Resolving the above equation, we get,
⇒a2−2a−a+2=0\Rightarrow {{a}^{2}}-2a-a+2=0
Taking the common factors, we get,
⇒a(a−2)−1(a−2)=0\Rightarrow a\left( a-2 \right)-1\left( a-2 \right)=0
The above equation can also be written as follows,
⇒(a−2)(a−1)=0\Rightarrow \left( a-2 \right)\left( a-1 \right)=0
Now,
⇒(a−2)=0 or (a−1)=0\Rightarrow \left( a-2 \right)=0\text{ }or\text{ }\left( a-1 \right)=0
⇒a=2 or a=1\Rightarrow a=2\text{ }or\text{ }a=1
∴\therefore The values of a are 1, 2 respectively.

Note: The reciprocal is a number that is multiplied by a given number and gives one as the product. It is like turning the number upside down or flipping it over. The reciprocal of the number xx is denoted by 1x\dfrac{1}{x} .