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Discriminant

Find 'k' it...

Question

Find $^{'} k^{'}$ its $(2 k + 1) x^{2} - 2 k x + k$ has real roots.

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Solution

If the quadratic equation $a x^{2} + b x + c = 0$ has real roots, then
$b^{2} - 4 a c ⩾ 0$

Given equation

$(2 k + 1) x^{2} - 2 k x + k$

Here $a \to 2 k + 1$
$b \to - 2 k$
$c \to k$

Substituting these values in the above inequality, we get

$\Rightarrow (- 2 k)^{2} - 4 (2 k + 1) k ⩾ 0$

$\Rightarrow 4 k^{2} - 8 k^{2} - 4 k ⩾ 0$

$\Rightarrow - 4 k^{2} - 4 k ⩾ 0$

$\Rightarrow - 4 (k^{2} + k) ⩾ 0$

$\Rightarrow (k^{2} + k) ⩽ 0$

$\Rightarrow k (k + 1) ⩽ 0$

Therefore, $\Rightarrow - 1 ⩽ k ⩽ 0$

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