The-values-of-k-for-which-the-equation-x-2-x-2-2k-1-1-k-2-has-repeated-roots-when-k-belongs-to-1-1-1-2-0-1-3-0-1-4-2-3-

Question Number 21071 by Tinkutara last updated on 12/Sep/17

The values of^{'} k^{'} for which the equation

∣ x ∣^{2} (∣ x ∣^{2} - 2 k + 1) = 1 - k^{2}, has

repeated roots, when k belongs to

(1) {1, - 1}

(2) {0, 1}

(3) {0, - 1}

(4) {2, 3}

Answered by mrW1 last updated on 13/Sep/17

let u =∣ x ∣^{2}

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

\Rightarrow u^{2} - (2 k - 1) u + (k^{2} - 1) = 0

\Rightarrow u = \frac{2 k - 1 \pm \sqrt{{(2 k - 1)}^{2} - 4 (k^{2} - 1)}}{2}

\Rightarrow u =∣ x ∣^{2} = \frac{2 k - 1 \pm \sqrt{5 - 4 k}}{2} ⩾ 0

5 - 4 k ⩾ 0

\Rightarrow k ⩽ \frac{5}{4} = 1.25 \dots (1)

2 k - 1 + \sqrt{5 - 4 k} ⩾ 0

\Rightarrow k ⩾ - 1 \dots (2)

2 k - 1 - \sqrt{5 - 4 k} ⩾ 0

\Rightarrow k ⩾ 1 \dots (3)

(1) + (2) + (3) \Rightarrow k \in [1, \frac{5}{4}]

? ? ?

Answered by Tinkutara last updated on 15/Sep/17

For repeated roots, ∣ x ∣ = 0 as for ∣ x ∣ \neq 0

roots will not be same .

\Rightarrow k^{2} = 1

k = \pm 1