Question

If the roots of the given equation $(2k+1){{x}^{2}}-(7k+3)x+k+2=0$ are reciprocal to each other, then the value of k will be
( a ) 0
( b ) 1
( c ) 2
( d ) 3

Answer 1

Hint: Here we are given the equation $(2k+1){{x}^{2}}-(7k+3)x+k+2=0$ and we have to find the value of k. As the roots are reciprocal to each other. So we suppose the roots to find the value and then by multiplying the roots and comparing that equation with the given equation, we are able to find the value of k.

Formula used:
Roots are reciprocal so we use that the product of roots will be equal to 1.

Complete step by step Solution:
Given equation is$(2k+1){{x}^{2}}-(7k+3)x+k+2=0$
We can write the above equation as ${{x}^{2}}-\dfrac{7k+3}{2k+1}x+\dfrac{k+2}{2k+1}=0$………………………………… (1)
We have to find the value of k
As the roots are reciprocal to each other.
So the product of the roots should be equal to 1.
Now we suppose that equation (1) has the roots $\alpha $ and $\dfrac{1}{\alpha }$, then
$(x-\alpha )\left( x-\dfrac{1}{\alpha } \right)=0$
That is ${{x}^{2}}-\left( \alpha \times \dfrac{1}{\alpha } \right)x+1=0$……………………………………………….. (2)
Now we compare the equation (1) and (2), and we get
$\alpha \times \dfrac{1}{\alpha }=\dfrac{7k+3}{2k+1}$ and $\dfrac{k+2}{2k+1}=1$
Solving the equations, we get
 2k + 1 = 7k + 3 and k + 2 = 2k + 1
-5k = 2 and k = 1
That is $k=-\dfrac{2}{5}$ and k = 1
As the negative sign is not possible, so
K = 1
So, the value of k = 1

Therefore, the correct option is (b).

Note:Quadratic equations are those polynomial equations that have a degree of 2. We should know that if the equations are reciprocal to each other then we have to take the roots on our own. We should take care while doing calculations.