Step-by-step explanation:
[tex] \green{\large\underline{\sf{Solution-}}}[/tex]
Given quadratic equation is
[tex]\rm :\longmapsto\:\rm \: {(3x)}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0 [/tex]
can be rewritten as
[tex]\rm :\longmapsto\:\rm \: {9x}^{2} + \bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 \bigg)x + 4 = 0 [/tex]
Concept Used :-
Nature of roots :-
Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of quadratic equation depends upon Discriminant (D) of the quadratic equation.
If Discriminant, D > 0, then roots of the equation are real and unequal.
If Discriminant, D = 0, then roots of the equation are real and equal.
If Discriminant, D < 0, then roots of the equation are unreal or complex or imaginary.
Where,
Discriminant, D = b² - 4ac
Let's Solve this problem now!!!
On comparing with quadratic equation ax² + bx + c = 0, we get
[tex]\red{\rm :\longmapsto\:a = 9}[/tex]
[tex]\red{\rm :\longmapsto\:b = 27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15}[/tex]
[tex]\red{\rm :\longmapsto\:c = 4}[/tex]
Since, Discriminant, D = 0
[tex]\rm \implies\: {b}^{2} - 4ac = 0[/tex]
[tex]\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 4 \times 4 \times 9 = 0[/tex]
[tex]\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} - 144 = 0[/tex]
[tex]\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = 144[/tex]
[tex]\rm :\longmapsto\: {\bigg(27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15\bigg)}^{2} = {12}^{2} [/tex]
[tex]\rm \implies\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = \: \pm \: 12[/tex]
Case - 1
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = - 12[/tex]
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = - 12 + 15[/tex]
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 3[/tex]
[tex]\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{9} [/tex]
[tex]\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = \dfrac{1}{ {3}^{2} } [/tex]
[tex]\rm \implies\:{\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{ - 2} [/tex]
[tex]\rm \implies\:\dfrac{1}{k} = - 2[/tex]
[tex]\bf\implies \:k \: = \: - \: \dfrac{1}{2} [/tex]
So, option (b) is Correct.
Case - 2
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } - 15 = 12[/tex]
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 12 + 15[/tex]
[tex]\rm :\longmapsto\:27 \times {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 27[/tex]
[tex]\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = 1[/tex]
[tex]\rm :\longmapsto\: {\bigg[3\bigg]}^{ \dfrac{1}{k} } = {3}^{0} [/tex]
[tex]\rm \implies\:\dfrac{1}{k} =0[/tex]
which is not possible.
