The-value-of-k-for-which-the-quadratic-equation-1-2k-x-2-6kx-1-0-and-kx-2-x-1-0-have-atleast-one-roots-in-common-are-

Question Number 97512 by bemath last updated on 08/Jun/20

The value of k for which the

quadratic equation (1 - 2 k) x^{2} - 6 kx - 1 = 0

and {kx}^{2} - x + 1 = 0 have atleast

one roots in common are___

Commented by john santu last updated on 08/Jun/20

Let common root be β .

(1 - 2 k) β^{2} - 6 k β - 1 = 0

k β^{2} - β + 1 = 0

\Rightarrow \frac{β^{2}}{- 6 k - 1} = \frac{β}{k - 1} = \frac{1}{{6 k}^{2} - (1 - 2 k)}

\Rightarrow {(k - 1)}^{2} = - (6 k + 1) ({6 k}^{2} + 2 k - 1)

{36 k}^{3} + {19 k}^{2} - 6 k = 0

k ({36 k}^{2} + 19 k - 6) = 0

k = 0 rejected

{36 k}^{2} + 19 k - 6 = 0

(4 k + 3) (9 k - 2) = 0

{\begin{cases} k = - \frac{3}{4} \\ k = \frac{2}{9} \end{cases}

Commented by bemath last updated on 08/Jun/20

thank you

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