Question

I am not getting the answer
pls help​

Answer 1

Answer:

Required value of k is 4.

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 }

= > k - 1 = 3

= > k = 4

Hence the required value of k is 4.

Answer 2

Question :--- one root of Equation (k-1)x² -10x +3 = 0 is the reciprocal of the other, then the value of k is ?

Formula used :--

→ sum of roots of Equation ax² + bx + c = 0 is given by = (-b/a)

→ Product of roots = (c/a) .

___________________________

Solution :---

Given Equation is (k-1)x² -10x +3 = 0 , where ,

→ a = (k-1)

→ b = (-10)

→ c = 3

Let roots of Equation are p and q.

it is given that now , roots are reciprocal .

so , we can say that ,

→ p = 1/q ------------ Equation (1)

Now, Above Told Formula , product of roots = c/a.

→ p * q = 3/(k-1)

Putting value of P from Equation (1) we get,

→ 1/q * q = 3/(k-1)

→ 1 = 3/(k-1)

→ k-1 = 3

→ k = 3+1

→ k = 4..

Hence, value of k will be 4 , when roots of the Equation are Reciprocal ...