Question

Solve it:–
$\sf{2x \: - \: 3 \: = \: \sqrt{2x {}^{2} - 2x + 21 } }$
_______
It can be solved by quadratic method.

Dont sPaM
​

Answer 1

Required answer :

The value of x is 6 for the given equation.

$~~~~~~~~$_____________

Explaination:

Given :

$\bf : \implies \: \: \red{ 2x - 3 = \sqrt{2 {x}^{2} - 2x + 21} }$

$\:$

Squaring on both sides :

$\bf : \implies \: \:\purple{ {(2x - 3)}^{2} = ( { \sqrt{2 {x}^{2} - 2x + 21} )}^{2}}$

$\:$

Using the identity (a – b)² = a² + b² – 2ab for LHS :

$\bf : \implies \: \: \red{ {(2x) ^{2} + {(3)}^{2} - 2(2x)(3) = {( \sqrt{2 {x}^{2} - 2x + 21 } )}^{2} }}$

$\:$

$\bf : \implies \: \: \purple{4 {x}^{2} - 12x + 9 = {( \sqrt{2 {x}^{2} - 2x + 21 } )}^{2} }$

$\:$

Square and square root gets cancelled on RHS :

$\bf : \implies \: \: \red{4 {x}^{2} - 12x + 9 = 2 {x}^{2} - 2x + 21 }$

$\:$

$\bf : \implies \: \: \purple{4 {x}^{2} - 2 {x}^{2} - 12x + 2x + 9 - 21 = 0 }$

$\:$

$\bf : \implies \: \: \red{2 {x}^{2} - 10x - 12 = 0 }$

$\:$

Removing 2 as the common term and cancelling :

$\bf : \implies \: \: \purple{2 \: ( {x}^{2} - 5x - 6) = 0 }$

$\:$

$\bf : \implies \: \: \red{ {x}^{2} - 5x - 6 = 0 \div 2 }$

$\:$

$\bf : \implies \: \: \purple{ {x}^{2} - 5x - 6 = 0 }$

$\:$

Factorising it by splitting the middle term :

$\bf : \implies \: \: \red{ {x}^{2} - 6x + x - 6 = 0 }$

$\:$

$\bf : \implies \: \: \purple{x \: (x - 6) + 1 \: (x - 6) = 0}$

$\:$

$\bf : \implies \: \: \red{(x - 6) \: (x + 1) = 0}$

$\:$

${ \bf : \implies \: \: \purple{ x = 6 \: and \: x = - 1}}$

$\:$

The value of x can't be – 1 as it can't be negative.

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Verification:

Firstly, substituting x with 6 :

$\bf : \implies \: \: \red{2x - 3 = \sqrt{2 {x}^{2} - 2x + 21} }$

$\:$

$\bf : \implies \: \: \purple{2(6) - 3 = \sqrt{2 {(6)}^{2} - 2(6) + 21 } }$

$\:$

$\bf : \implies \: \: \red{12 - 3 = \sqrt{2(36) - 12 + 21} }$

$\:$

$\bf : \implies \: \: \purple{9 = \sqrt{72 + 9} }$

$\:$

$\bf : \implies \: \: \red{9 = \sqrt{81} }$

$\:$

$\bf : \implies \: \: \purple{9 = \: 9}$

$~$

$\bf :\implies~~ \red{LHS=RHS}$

$\:$

Henceforth, verified!

Answer 2

$\begin{gathered} \\ \large{ \underline{ \underline{ \pmb{ \sf{ \maltese{ \color{red}{ \: Given \: in \: the \: Question \: that, \: }}}}}}} \\ \\ \end{gathered}$

$\begin{gathered} \\ \sf{2x \: - \: 3 \: = \: \sqrt{2x {}^{2} - 2x + 21 } } \\ \\ \end{gathered}$

$\begin{gathered} \\ \large{ \underline{ \underline{ \pmb{ \sf{ \maltese{ \color{blue}{ \: On \: Squaring \: both \: the \: sides, \: }}}}}}} \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{(2x \: - \: 3) {}^{2} \: = \: {2x {}^{2} - 2x + 21 } } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{(2x) {}^{2} \: - \: (2) \: (2x) \: (3) \: = \: {2x {}^{2} - 2x + 21 } } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{(4x) {}^{2} + 9 - 12x \: = \: {2x {}^{2} - 2x + 21 } } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{2x ^{2} = 10x + 12 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{2x ^{2} - 10x - 12 = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \large{ \underline{ \underline{ \pmb{ \sf{ \maltese{ \color{springgreen}{ \: On \: Splitting \: the \: mid \: term, \: }}}}}}} \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{2x ^{2} - 12x + 2x - 12 = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{2x(x - 6) + (x - 6) = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{(2x + 2)(x - 6) = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \large{ \underline{ \underline{ \pmb{ \sf{ \maltese{ \color{purple}{ \: Therefore \: the \: values \: of \: x \: are, \: }}}}}}} \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{(2x + 2) = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{x = - 1 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{ (x - 6) = 0 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \: \longmapsto \: \sf{ x = 6 } \\ \\ \end{gathered}$

$\begin{gathered} \\ \\ \underline{\rule{250pt}{8pt}} \\ \end{gathered}$

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