Question

If one zeroes of the quadratic polynomial (k-1)x²-10x+3is the reciprocal of the other, then the value of K is------​

Answer 1

Step-by-step explanation:

Let the roots of the quadratic polynomial be x and 1/x, then

: (k-1)x² - 10x + 3 = 0

: x = 10±√(100 - 12k + 12)/2k - 2

: x = 10+√(100-12k+12)/2k - 2 , 1/x = 10-√100-12k+12/2k - 2

Equation formed further :

:

$\frac{10 + \sqrt{100 - 12x + 12} }{2k - 2} = \frac{2k - 2}{10 - \sqrt{100 - 12k + 12k} }$

:

${(2k - 2)}^{2} = {10}^{2} - ( { \sqrt{100 - 12k + 12} })^{2}$

:

$( {2k - 2)}^{2} = 12k - 12$

:

${4k}^{2} - 8k + 4 = 12k - 12$

:

${4k}^{2} - 20k + 16 = 0$

:

${k}^{2} - 5k + 4 = 0$

:

$(k - 1)(k - 4) = 0$

:

$k = 1 and 4$

Since if we take 'k' as 1 the equation won't work so Hence, k = 4

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