Answer :
Answer:
[tex]\boxed{\bf\: k \: \in \: ( - 3, \: 3) \: } \\ [/tex]
Step-by-step explanation:
Given that, the quadratic equations kx² - 6x + k = 0 has real and distinct roots.
On comparing with general quadratic equation ax² + bx + c = 0, we get a = k, b = - 6, c = k
As it is given that, the quadratic equations kx² - 6x + k = 0 has real and distinct roots, so it means,
[tex]\sf\:Discriminant, D > 0\\[/tex]
[tex]\sf\: {b}^{2} - 4ac > 0 \\ [/tex]
[tex]\sf\: {( - 6)}^{2} - 4 \times k \times k > 0 \\ [/tex]
[tex]\sf\: 36 - 4 {k}^{2} > 0 \\ [/tex]
[tex]\sf\: - 4 ({k}^{2} - 9) > 0 \\ [/tex]
[tex]\sf\: {k}^{2} - 9 < 0 \\ [/tex]
[tex]\sf\: {k}^{2} - {3}^{2} < 0 \\ [/tex]
[tex]\sf\: (k + 3)(k - 3) < 0 \\ [/tex]
[tex]\implies\sf\: - 3 < k < 3 \\ [/tex]
[tex]\implies\sf\: k \: \in \: ( - 3, \: 3) \\ [/tex]
Hence,
[tex]\implies\sf\:\boxed{\bf\: k \: \in \: ( - 3, \: 3) \: } \\ [/tex]
[tex]\rule{190pt}{2pt}[/tex]
Additional Information:
[tex]\:\boxed{\begin{aligned}& \:\sf \:(x - a)(x - b) < 0, \: \: \sf\implies \: x \: \in \: (a, \: b), \: \: where \: a < b \: \\ \\& \:\sf \:(x - a)(x - b) \leqslant 0, \: \: \sf\implies \: x \: \in \: [ a, \: b], \: \: where \: a < b \\ \\& \:\sf \:(x - a)(x - b) > 0, \: \sf\implies x \: \in \: ( - \infty , \: a) \cup \: (b, \infty ), \: \: where \: a < b\\ \\& \:\sf \:(x - a)(x - b) \geqslant 0, \: \sf\implies x \: \in \: ( - \infty , \: a] \cup \: [ b, \infty ), \: \: where \: a < b\end{aligned}}[/tex]