Question

If one root of the equation (k-1)x^2-10x+3=0 is the reciprocal of the other ,then the value of k is ______.

Answer 1

Answer:

k = 4

Step-by-step explanation:

Let say One root is

P then other root is 1/P

Products of roots = c/a

=> P * (1/P) = c/a

=> 1 = c/a

=> c = a

=> k - 1 = 3

=> k = 4

(4-1)x² - 10x + 3 = 0

=> 3x² - 9x - x + 3 = 0

=> 3x(x - 3) - 1(x - 3) = 0

=>(3x - 1)(x - 3) = 0

=> x = 1/3 , 3

Another method

roots P & 1/P

(x -  P)(x - 1/P) = 0

=> x² - x(P + 1/P)  + 1 = 0

multiplying by 3

=> 3x² - 3x(P + 1/P) + 3 = 0

Equating with

(k-1)x² - 10x + 3 = 0

=> k - 1 = 3 => k = 4

3(P + 1/P) = 10

=> 3P² - 10P + 3 = 0

=> (3P - 1)(P -3) = 0

P = 1/3 , 3

Answer 2

Answer:

The value of k is 4.

Given Data:

Equation:

$(K-1) x^{2}-10 x+3=0$

Step 1:

$(K-1) x^2-10x +3=0$

Let the two roots are s and \frac{1}{s}

$\mathbf{s} \times \frac{\mathbf{1}}{s}=\frac{c}{s}$

$\mathbf{s} \times \frac{\mathbf{1}}{s}=\frac{\mathbf{3}}{k-1}$

Step 2:

Evaluate the LHS expression.

$\mathrm{s} \times \frac{1}{s}=1$

Substitute this in the LHS.

Step 3:

1$=\frac{3}{k-1}$

Bring k-1 to the LHS.

K-1 = 3

Step 4:

Bring 1 to the RHS

K=3+1 (add the RHS value, so we will get 4)

The value of K=4