Given :
[tex] \\ \\ [/tex]
To Find :
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
SolutioN :
[tex] \dag \; {\underline{\underline{\sf{ \; Applying \; the \; Value \; :- }}}} [/tex]
[tex] \; \; \implies \; \; \sf { {x}^{2} + k(x) - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + k \bigg( \dfrac{1}{2} \bigg) - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k = 0 + \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k = \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{4} + \dfrac{1}{2}k = \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{5}{4} - \dfrac{1}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{5 - 1}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{4}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \cancel\dfrac{4}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = 1 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { 1k = 1 \times 2 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { 1k = 2 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; {\pmb{\underline{\boxed{\red{\frak { k = \dfrac{2}{1} }}}}}} \; \purple\bigstar \\ \\ \\ [/tex]
[tex] \qquad \; \therefore \; [/tex] Value of k is 2/1
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Answers & Comments
Given :
[tex] \\ \\ [/tex]
To Find :
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
SolutioN :
[tex] \dag \; {\underline{\underline{\sf{ \; Applying \; the \; Value \; :- }}}} [/tex]
[tex] \; \; \implies \; \; \sf { {x}^{2} + k(x) - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + k \bigg( \dfrac{1}{2} \bigg) - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k - \dfrac{5}{4} = 0 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k = 0 + \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { { \bigg( \dfrac{1}{2} \bigg) }^{2} + \dfrac{1}{2}k = \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{4} + \dfrac{1}{2}k = \dfrac{5}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{5}{4} - \dfrac{1}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{5 - 1}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \dfrac{4}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = \cancel\dfrac{4}{4} } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { \dfrac{1}{2}k = 1 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { 1k = 1 \times 2 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; \sf { 1k = 2 } \\ \\ \\ [/tex]
[tex] \; \; \implies \; \; {\pmb{\underline{\boxed{\red{\frak { k = \dfrac{2}{1} }}}}}} \; \purple\bigstar \\ \\ \\ [/tex]
[tex] \\ \\ [/tex]
[tex] \qquad \; \therefore \; [/tex] Value of k is 2/1
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]