Question

If one zero of the polynomial \[p(x)=5{{x}^{2}}+13x-a\] is the reciprocal of the other, find the value of a.
A. 5
B. -5
C. 7
D. -7

Answer 1

Hint: As we have one zero of a polynomial which is reciprocal to the other. So we will use the formula for products of roots because in this product of roots will be 1.

Complete step-by-step solution -
So if we have polynomial $p(x)=A{{x}^{2}}+Bx+C$ then the product of roots equal to $\dfrac{C}{A}$ .
Since given that the other zero is the reciprocal of the one zero.
The given polynomial function is \[p(x)=5\mathop{x}^{2}+13x-a\]
Let us suppose that \[\alpha \] is one zero of the polynomial and \[\dfrac{1}{\alpha }\] is another zero.
On comparing given polynomial with $p(x)=A{{x}^{2}}+Bx+C$
$A=5,B=13,C=-a$
As we have the product of roots equal to $\dfrac{C}{A}$
So we can write
$\Rightarrow \alpha \times \dfrac{1}{\alpha }=\dfrac{C}{A}$
$\Rightarrow 1=\dfrac{-a}{5}$
$\Rightarrow a=-5$
Hence we get the required value of a is -5
Therefore, option (B) is the right answer.

Note: In this, we need to be careful about the reciprocal of others. To find reciprocal of any number we divide 1 by that number. If we have a number that is 2 then reciprocal of 2 is $\dfrac{1}{2}$. Hence $\Rightarrow 2\times \dfrac{1}{2}=1$. Multiplication of reciprocal of number and that number is 1.