If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is

1.	If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is :-a) rootb) not a rootc) sum of the rootsd) product of the roots
Answer» $\large\underline{\sf{Solution-}}$ Given that $\red{\rm :\longmapsto\: \alpha \:is \: the \: root \: of \: {4x}^{2} = 2x - <klux>1</klux>}$ can be rewritten as $\red{\rm :\longmapsto\: \alpha \:is \: the \: root \: of \: {4x}^{2} - 2x + 1 = 0}$ $\bf\implies \: {4\alpha \:}^{2} - 2\alpha \: + 1 = 0$ or $\bf\implies \: {4\alpha \:}^{2} = 2\alpha \: - 1 - - - (1)$ Now, Consider, $\red{\rm :\longmapsto\: {4\alpha \:}^{3} - 3 \alpha }$ can be rewritten as $\rm \: = \: \: \alpha \:( {4\alpha \:}^{2} ) - 3\alpha \:$ On Substituting the value from equation (1), we get $\rm \: = \: \: \alpha \:( 2\alpha \: - 1 ) - 3\alpha \:$ $\rm \: = \: \: {2\alpha \:}^{2} -\alpha \: - 3\alpha \:$ $\rm \: = \: \: {2\alpha \:}^{2} -4\alpha \:$ $\rm \: = \: \:\dfrac{1}{2} \bigg( {4\alpha \:}^{2} -8\alpha \: \bigg)$ On substituting the value from equation (1), we get $\rm \: = \: \:\dfrac{1}{2} \bigg( 2 \alpha - 1 -8\alpha \: \bigg)$ $\rm \: = \: \:\dfrac{1}{2} \bigg( - 1 -6\alpha \: \bigg)$ $\rm \: = \: \: - \: \dfrac{1}{2} - 3\alpha \:$ Now, It implies, $\red{\rm :\longmapsto\: {4\alpha \:}^{3} - 3 \alpha = - \: \dfrac{1}{2} - 3 \alpha }$ LET we CHECK, whether its a root or not. So, Let we substitute the above value in $\red{\rm :\longmapsto\: {4x}^{2} = 2x - 1}$ $\rm :\longmapsto\:4 {\bigg( - \dfrac{1}{2} - 3\alpha \:\bigg) }^{2} = 2\bigg( - \dfrac{1}{2} - 3 \alpha \bigg) - 1$ $\rm :\longmapsto\:4 {\bigg( - \dfrac{(1 + 6\alpha \:)}{2} \:\bigg) }^{2} = 2\bigg( - \dfrac{1 + 6\alpha \:}{2} \bigg) - 1$ $\red{\rm :\longmapsto\: {(1 + 6\alpha \:)}^{2} = - (1 + 6\alpha \:) - 1}$ $\rm :\longmapsto\:1 + {36\alpha \:}^{2} + 12\alpha \: = - 1 - 6\alpha \: - 1$ $\rm :\longmapsto\: {36\alpha \:}^{2} = - 3 - 18\alpha$ $\rm :\longmapsto\: {12\alpha \:}^{2} = - 1 - 6\alpha$ $\rm :\longmapsto\: 3 {(4\alpha \:}^{2}) = - 1 - 6\alpha$ $\rm :\longmapsto\: 3 (2 \alpha - 1) = - 1 - 6\alpha$ $\rm :\longmapsto\: 6 \alpha -3 = - 1 - 6\alpha$ $\bf\implies \: {4\alpha \:}^{3} - 3\alpha \: \: is \: not \: the \: root \: of \: given \: equation.$ Now, Let we check its SUM of the roots. We know, $\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$ $\red{\rm :\longmapsto\:\alpha \: - \dfrac{1}{2} - 3\alpha \: = - \dfrac{( - 2)}{4}}$ $\red{\rm :\longmapsto\: \: - \dfrac{1}{2} - 2\alpha \: = \dfrac{1}{2}}$ which is not true. Now, Check the PRODUCT of the roots We know, $\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$ $\red{\rm :\longmapsto\:\alpha \: \times \bigg( - \dfrac{1}{2} - 3 \alpha \bigg) = \dfrac{1}{4}}$ Which is not true Hence, Option (b) is correct.

If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is :-a) rootb) not a rootc) sum of the rootsd) product of the roots

Answer»

$\large\underline{\sf{Solution-}}$

Given that

$\red{\rm :\longmapsto\: \alpha \:is \: the \: root \: of \: {4x}^{2} = 2x - <klux>1</klux>}$

can be rewritten as

$\red{\rm :\longmapsto\: \alpha \:is \: the \: root \: of \: {4x}^{2} - 2x + 1 = 0}$

$\bf\implies \: {4\alpha \:}^{2} - 2\alpha \: + 1 = 0$

$\bf\implies \: {4\alpha \:}^{2} = 2\alpha \: - 1 - - - (1)$

Now,

Consider,

$\red{\rm :\longmapsto\: {4\alpha \:}^{3} - 3 \alpha }$

can be rewritten as

$\rm \: = \: \: \alpha \:( {4\alpha \:}^{2} ) - 3\alpha \:$

On Substituting the value from equation (1), we get

$\rm \: = \: \: \alpha \:( 2\alpha \: - 1 ) - 3\alpha \:$

$\rm \: = \: \: {2\alpha \:}^{2} -\alpha \: - 3\alpha \:$

$\rm \: = \: \: {2\alpha \:}^{2} -4\alpha \:$

$\rm \: = \: \:\dfrac{1}{2} \bigg( {4\alpha \:}^{2} -8\alpha \: \bigg)$

On substituting the value from equation (1), we get

$\rm \: = \: \:\dfrac{1}{2} \bigg( 2 \alpha - 1 -8\alpha \: \bigg)$

$\rm \: = \: \:\dfrac{1}{2} \bigg( - 1 -6\alpha \: \bigg)$

$\rm \: = \: \: - \: \dfrac{1}{2} - 3\alpha \:$

Now, It implies,

$\red{\rm :\longmapsto\: {4\alpha \:}^{3} - 3 \alpha = - \: \dfrac{1}{2} - 3 \alpha }$

LET we CHECK, whether its a root or not.

So, Let we substitute the above value in

$\red{\rm :\longmapsto\: {4x}^{2} = 2x - 1}$

$\rm :\longmapsto\:4 {\bigg( - \dfrac{1}{2} - 3\alpha \:\bigg) }^{2} = 2\bigg( - \dfrac{1}{2} - 3 \alpha \bigg) - 1$

$\rm :\longmapsto\:4 {\bigg( - \dfrac{(1 + 6\alpha \:)}{2} \:\bigg) }^{2} = 2\bigg( - \dfrac{1 + 6\alpha \:}{2} \bigg) - 1$

$\red{\rm :\longmapsto\: {(1 + 6\alpha \:)}^{2} = - (1 + 6\alpha \:) - 1}$

$\rm :\longmapsto\:1 + {36\alpha \:}^{2} + 12\alpha \: = - 1 - 6\alpha \: - 1$

$\rm :\longmapsto\: {36\alpha \:}^{2} = - 3 - 18\alpha$

$\rm :\longmapsto\: {12\alpha \:}^{2} = - 1 - 6\alpha$

$\rm :\longmapsto\: 3 {(4\alpha \:}^{2}) = - 1 - 6\alpha$

$\rm :\longmapsto\: 3 (2 \alpha - 1) = - 1 - 6\alpha$

$\rm :\longmapsto\: 6 \alpha -3 = - 1 - 6\alpha$

$\bf\implies \: {4\alpha \:}^{3} - 3\alpha \: \: is \: not \: the \: root \: of \: given \: equation.$

Now, Let we check its SUM of the roots.

We know,

$\boxed{\red{\sf Sum\ of\ the\ zeroes=\frac{-coefficient\ of\ x}{coefficient\ of\ x^{2}}}}$

$\red{\rm :\longmapsto\:\alpha \: - \dfrac{1}{2} - 3\alpha \: = - \dfrac{( - 2)}{4}}$

$\red{\rm :\longmapsto\: \: - \dfrac{1}{2} - 2\alpha \: = \dfrac{1}{2}}$

which is not true.

Now, Check the PRODUCT of the roots

We know,

$\boxed{\red{\sf Product\ of\ the\ zeroes=\frac{Constant}{coefficient\ of\ x^{2}}}}$

$\red{\rm :\longmapsto\:\alpha \: \times \bigg( - \dfrac{1}{2} - 3 \alpha \bigg) = \dfrac{1}{4}}$

If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is :-a) rootb) not a rootc) sum of the rootsd) product of the roots

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If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is :-a) rootb) not a rootc) sum of the rootsd) product of the roots​

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If 'α' is a root of the equation 4x² = 2x - 1 then '4α³ - 3α' is :-a) rootb) not a rootc) sum of the rootsd) product of the roots