Question

find the value of k

→ stu*pid mods, don't delete my question​

Answer 1

Answer:

In each of the following, find the value of k for which the given value is a solution of the given equation :

So, here we are given some equations, which contains two variables out of which the value of ' x 'is provided accordingly and we have to find the actual value of other variable i.e, k in each equation. For that, firstly we will substitute the value of ' x ' and thereafter by taking the equation equal to zero we will be able to find the value of the ' k '

So, Without any further delay, let's do it :-

$\implies \sf i) \: 7x^{2} + kx - 3 = 0$

Given :

$\dashrightarrow\:\:\bf x = \dfrac{2}{3}$

Plugging in the value of ' x ' in the above equation :-

$\dashrightarrow\:\:\sf 7 \times {\red{\dfrac{2}{3}^{2}}} + k \times \dfrac{2}{3} - 3 = 0$

$\dashrightarrow\:\:\sf 7 \times \dfrac{4}{9} + \dfrac{2}{3}k - 3 = 0$

$\dashrightarrow\:\:\sf \dfrac{28}{9} + \dfrac{2}{3}k - 3 = 0$

$\dashrightarrow\:\:\sf \dfrac{28 + 6k - 27}{9} = 0$

$\dashrightarrow\:\:\sf \dfrac{1 + 6k }{9} = 0$

$\dashrightarrow\:\:\sf 1 + 6k = 0$

$\dashrightarrow\:\:\sf 6k = - 1$

$\dashrightarrow\:\:\sf k = \dfrac{-1}{6}$

Hence

${\boxed{\bf {The \: value \: of \: k = {\purple{\dfrac{-1}{6} }}}}}$

____________________

$\implies \sf \: 2) \: x^{2} - x(a + b) + k = 0$

Given :

$\dashrightarrow\:\:\bf x = a$

Plugging in the value of ' x ' in the above equation :

$\dashrightarrow\:\:\sf a^{2} - a(a + b) + k = 0$

$\dashrightarrow\:\:\sf a^{2} - a^{2} - ab + k = 0$

$\dashrightarrow\:\:\sf {\not{a^{2}}} - {\not{a^{2}}} - ab + k = 0$

$\dashrightarrow\:\:\sf - ab + k = 0$

$\dashrightarrow\:\:\sf k = ab$

${\boxed{\bf {The \: value \: of \: k = {\purple{ab}}}}}$

__________________

$\implies \sf \: 3) \: kx^{2} + \sqrt 2 x - 4 = 0$

Given :

$\dashrightarrow\:\:\bf x = \sqrt 2$

Plugging in the value of ' x ' in the above equation :-

$\dashrightarrow\:\:\sf k ({\sqrt 2})^{2} + \sqrt 2 (\sqrt 2) - 4 = 0$

$\dashrightarrow\:\:\sf 2k + 2 - 4 = 0$

$\dashrightarrow\:\:\sf 2k - 2 = 0$

$\dashrightarrow\:\:\sf 2k = 2$

$\dashrightarrow\:\:\sf k = \dfrac{\not 2}{\not 2^{ \: \: 1 }}$

${\boxed{\bf {The \: value \: of \: k = {\purple{1} }}}}$

__________________

$\implies \sf 4) \: x^{2} + 3ax + k = 0$

Given :

$\dashrightarrow\:\:\bf x = (- a)$

Plugging in the value of ' x ' in the above equation :-

$\dashrightarrow\:\:\sf (- a)^{2} + 3a(- a) + k = 0$

$\dashrightarrow\:\:\sf a^{2} - 3a^{2} + k = 0$

$\dashrightarrow\:\:\sf - 2a^{2} + k = 0$

$\dashrightarrow\:\:\sf k = 2a^{2}$

${\boxed{\bf {The \: value \: of \: k = {\purple{2a^{2}}}}}}$

__________________

Not sure about the answer?

Let's verify it :

For verification, we will simply plug in the value of the k in the equation, if it will satisfy and make the whole equation zero then it is correct answer.

1) $\longrightarrow\:\:\sf 1 + 6k = 0$

Putting x = $\sf \dfrac{-1}{6}$

$\rightarrow\:\:\sf 1 + 6 {\blue{\bigg(\dfrac{-1}{6}\bigg)}} = 0$

$\rightarrow\:\:\sf 1 + {\dfrac{-6}{6}} = 0$

Taking L.C.M

$\rightarrow\:\:\sf {\dfrac{6 -6}{6}} = 0$

$\rightarrow\:\:\sf {\dfrac{0}{6}} = 0$

$\rightarrow\:\:\sf 0 = 0$

${\underline{\bf{L. H. S = R. H. S}}}$

$\dag \: \: {\large {\sf{ Verified } \: \: \dag}}$

___________________

$\longrightarrow\:\:\sf \: 2) \: - ab + k = 0$

Putting k = ab

$\rightarrow\:\:\sf - ab + {\blue{(ab)}} = 0$

$\rightarrow\:\:\sf - {\not ab} + {\not{ab}} = 0$

$\rightarrow\:\:\sf 0 = 0$

${\underline{\bf{L. H. S = R. H. S}}}$

$\dag \: \: {\large {\sf{ Verified } \: \: \dag}}$

___________________

$\longrightarrow\:\:\sf \: 3) \: 2k - 2 = 0$

Putting K = 1

$\rightarrow\:\:\sf 2{\blue{(1)}} - 2 = 0$

$\rightarrow\:\:\sf 2 - 2 = 0$

$\rightarrow\:\:\sf 0 = 0$

${\underline{\bf{L. H. S = R. H. S}}}$

$\dag \: \: {\large {\sf{ Verified } \: \: \dag}}$

____________________

$\longrightarrow\:\:\sf \: 4) \: - 2a^{2} + k = 0$

Putting $\sf k = 2a^{2}$

$\rightarrow\:\:\sf - 2a^{2} + {\blue{(2a)^{2}}} = 0$

$\rightarrow\:\:\sf 0 = 0$

${\underline{\bf{L. H. S = R. H. S}}}$

$\dag \: \: {\large {\sf{ Verified } \: \: \dag}}$

____________________

Answer 2

SOLUTION :-

$\:$

$\small\mathtt\red{(I)}$$\textbf{Given\:that:-}$

$\:\:$

$\sf :\implies x\:=\:\bf\dfrac{2}{3}\:\:\:\:is \:a\:root\:of\:the\: equation$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\: x\:=\:\bf\dfrac{2}{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\sf |satisfies\:the\:Eqn|$

$\:\:$

$\sf \:\:\:\:\:\:\:\dashrightarrow i.e.\:\:7\bigg(\bf\dfrac{2}{3}\bigg)^2\:+\:k\bigg(\bf\dfrac{2}{3}\bigg)\:-\:3=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mapsto\:7×\bf\dfrac{4}{9}\:+\:2\bf\dfrac{k}{3}\:-\:3=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leadsto\:2\bf\dfrac{k}{3}\:=\:3-\bf\dfrac{28}{9}$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leadsto\:2 \bf\dfrac{k}{3}\:=\: \bf\dfrac{27-28}{9}$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leadsto\:2 \dfrac{k}{ \not{3}} = -\bf \dfrac{1}{\cancel{9}^{ \: \: 3} }$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{k=\frak{\green{- \bf\dfrac{1}{6}}}}}}$

$\:\:\:\:\:\:\:\:\:\:____________________$

$\:\:$

$\small\mathtt\red{(ii)}$$\textbf{Given\:that:-}$

$\:$

$\sf :\implies \bf x=a\:\:\:\:is\:a\:root\:of\:the\:given\: eqn.$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:x²-x(a+b)+k\:=\:0$

$\:\:$

.$\sf \:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\: x\:=\:a\:\:\:\:\:\:\:\:\:\:|satisfies\:the\:Eqn|$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\mapsto\: i.e.\:(a)²\:-\:a(a+b)+=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\leadsto\: \:(a)²\:-\:a²\:-\:ab\:+\:=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leadsto\:-ab\:+\:k=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{k=\frak{\orange{ab}}}}}$

$\:\:\:\:\:\:\:\:\:\:____________________$

$\:\:$

$\small\mathtt\red{(iii)}$$\textbf{Given\:that:-}$

$\:$

$\sf :\implies\:x=\bf\sf \displaystyle\sqrt{2}\bf\:is\:a\:root\:of\:the\:given\: eqn.$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:kx²+\bf\sf \displaystyle\sqrt{2}x\:-\:4\:=\:0$

$\:\:$

.$\sf \:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\: x\:=\:\bf\sf \displaystyle\sqrt{2}\:\:\:\:\:\:\:|satisfies\:the\:Eqn|$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\mapsto\: i.e.\:\:k(a\bf\sf \displaystyle\sqrt{2})²\:+\:\bf\sf \displaystyle\sqrt{2} (\bf\sf \displaystyle\sqrt{2})-4\:=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mapsto \:2k\:+\:2\:-\:4\:=0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \leadsto\:2k\:=\:2$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{k=\frak{\pink{1}}}}}$

$\:\:\:\:\:\:\:\:\:\:____________________$

$\:\:$

$\small\mathtt\red{(iv)}$$\textbf{Here,\:we\: have}$

$\:$

$\sf :\implies x\:=\:-0\:is\:a\:root\:of\:the\:given\: equation$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:x³+\:3ax\:+\:k\:=\:0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\dashrightarrow\: x\:=\:-a\:\:\:\:\:\:\:|satisfies\:the\:Eqn|$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\leadsto\: i.e.\:\:(-a)²\:+\:3a(-a)\:+\:k\:=\:0$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\leadsto\: -2a²\:+\:k\:=\:0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\boxed{\sf{k=\frak{\gray{2a²}}}}}$