Question

Find the value of k for which the root of equation 3xsquare - 10x +k=0 are reciprocal of each other

Answer 1

Answer:

Required value of k is 3.

Step-by-step explanation:

In a general quadratic equation ax^2 + bx + c = 0, values of x are given by using :

• $\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Given that the roots are reciprocal of each other :

$\implies\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{1}{\dfrac{-b-\sqrt{b^2-4ac}}{2a}}$

$\implies\dfrac{-b+\sqrt{b^2-4ac}}{2a}=\dfrac{2a}{-b-\sqrt{b^2-4ac}}$

= > ( 2a )^2 = ( - b )^2 - ( √( b^2 - 4ac ) )^2

= > 4a^2 = b^2 - b^2 + 4ac

= > 4a^2 = 4ac

= > a = c { c = constant term }

= > 3 = k

Hence the required value of k is 3.

Answer 2

Question :-- Find the value of k for which the root of equation 3xsquare - 10x +k=0 are reciprocal of each other ?

Concept and Formula used :--

→ For Quadratic Equation ax² + bx + c = 0 ,

=> sum of roots is = (-b/a)

=> Product of roots is = (c/a)

__________________________

Solution :--

Let roots of Equation 3x² -10x +k = 0 are α and β ...

As told above ,

→ α + β = -b/a = -(-10)/1 = 10 ----- Equation (1)

→ α × β = c/a = k/3 ------ Equation (2)

Now, it is given that, roots are reciprocal of each other ,

so,

→ β = 1/α ------------- Equation (3).

___________________________

Putting value of Equation (3) in Equation (2) now , we get,

→ α × 1/α = k/3

→ 1 = k/3

Cross - Multiply we get,

→ k = 3 .

Hence, value of k will be 3 ..