Question

If
$x = \frac{1}{2}$
is a root of the equation
${x}^{2} + kx - \frac{5}{4} = 0$

then find the value of k. ​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt If \:x = \dfrac{1}{2}\: is \:a \:root \:of \:the\:\\\tt equation\: {x}^{2} + kx - \frac{5}{4} =0.\:\\\tt Then\: find\: the\: value\: of\: k.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf\implies {x}^{2} + kx - \frac{5}{4} = 0$

$\sf \bf\implies x=\dfrac{1}{2}$

${\overbrace {\underbrace {\color {orange} {\sf \bf Substitute \:the \:value\:of \:x}}}}$

$\sf \bf\implies {\bigg(\dfrac{1}{2}\bigg)}^{2} + k\bigg(\dfrac{1}{2}\bigg) - \frac{5}{4} = 0$

$\sf \bf\implies \dfrac{1}{4}+\dfrac{1k}{2}-\dfrac{5}{4}=0$

${\color {orange} \underline {\tt By\:taking \:LCM}}$

$\sf \bf\implies \dfrac{1+2k-5}{4}=0$

$\sf \bf\implies 1+2k-5=0 \times 4$

$\sf \bf\implies 1+2k-5=0$

$\sf \bf\implies 2k-4=0$

$\sf \bf\implies 2k=4$

$\sf \bf\implies k= \dfrac{4}{2}$

$\implies{\boxed {\boxed {\color {blue} {\sf \bf k=2}}}}$

${\color {purple} \underline {\tt \therefore The\:value \:of \:k\:is\:2}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU }}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

${\color {green} {\tt Identities}}$

$\sf \bf {(a+b)}^{2}={a}^{2}+2ab+{b}^{2}$

$\sf \bf {(a-b)}^{2}={a}^{2}-2ab+{b}^{2}$

$\sf \bf (a+b) (a-b) ={a}^{2}-{b}^{2}$

Answer 2

Answer:

x + kx - = 0 4

1

- 2

Substitute the value of x

2 +k 5 4 2 2

1 4 + 1k 2 5 4 = 0

By taking LCM

1 + 2k - 5 = 0

4

1 + 2k - 5 = 0 x 4

1 + 2k - 5 = 0

2k - 4 = 0

+ 2k = 4

4

2

Step-by-step explanation:

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