Question

Solve quadratic equation

$a {}^{2} b {}^{2} x {}^{2} + b {}^{2}x - a {}^{2} x - 1 = 0$

Answer 1

$\underline{\underline{\large{\mathfrak{Solution : }}}}$



$\underline{\mathsf{To \: Solve \longrightarrow a^{2}b^{2}x^{2} \: + \: b^{2}x \: - \: a^{2}x \: - \: 1 \: = \: 0}}$



$\underline{\textsf{By Factorization Method : }} \\ \\ \mathsf{\implies a^{2}b^{2}x^{2} \: + \: b^{2}x \: - \: a^{2}x \: - \: 1 \: = \: 0 } \\ \\ \mathsf{\implies b^{2}x ( a^{2}x \: + \: 1 ) \: - \: 1( a^{2}x \: + \: 1 ) \: = \: 0} \\ \\ \mathsf{\implies (a^{2}x \: + \: 1)(b^{2}x \: - \: 1 ) \: = \: 0 } \\ \\ \mathsf{By \: Zero \: Product \: Rule : } \\ \\ \mathsf{\implies a^{2}x \: + \: 1 \: = \: 0 \quad \: or \quad b^{2}x \: - \: 1 \: = \: 0 } \\ \\ \mathsf{\implies a^{2}x \: = \: -1 \quad \: or \quad b^{2}x \: = \: 1 } \\ \\ \mathsf{\therefore \quad x \: = \: \dfrac{-1}{a^{2}} \quad \: or \quad x \: = \: \dfrac{1}{b^{2}}}$




$\underline{\textsf{By Quadratic Formula :}} \\ \\ \mathsf{\implies a^{2}b^{2}x^{2} \: + \: b^{2}x \: - \: a^{2}x \: - \: 1 \: = \: 0 } \\ \\ \mathsf{\implies (ab)^{2}x^{2} \: + \: ( b^{2} \: - \: a^{2})x \: - \: 1 \: = \: 0 }$

$\textsf{Here : } \\ \\ \mathsf{\implies Coefficient \: of \: x^{2} (\textbf{a}) \: = \: (ab)^{2}} \\ \\ \mathsf{\implies Coefficient \: of \: b( \bold{b}) \: = \: (b^{2} \: - \: a^{2} )} \\ \\ \mathsf{\implies Constant \: term (\textbf{c}) \: = \: - 1 }$




$\textsf{Using Quadratic Formula : } \\ \\ \mathsf{\implies x \: = \: \dfrac{- \bold{b} \: \pm \: \sqrt{ \bold{b}^{2} \: - \: 4 \bold{ac }}}{2 \bold{a}}}$

$\mathsf{\implies x \: = \: \dfrac{-(b^{2} \: - \: a^{2}) \: \pm \: \sqrt{(b^{2} \: - \: a^{2})^{2}\: - \: 4(ab)^{2} \: \times \: ( - 1)}}{2(ab)^{2}}}$



$\mathsf{\implies x \: = \: \dfrac{-b^{2} \: + \: a^{2} \: \pm \: \sqrt{b^{4} \: + \: a^{4} \: - \: 2 {a}^{2} {b}^{2} \: + \: 4 {a}^{2} {b}^{2} }}{2 {a}^{2} {b}^{2} }}$




$\mathsf{\implies x \: = \: \dfrac{-b^{2} \: + \: a^{2} \: \pm \: \sqrt{b^{4} \: + \: a^{4} \: + \: 2 {a}^{2} {b}^{2} \: }}{2 {a}^{2} {b}^{2} }}$




$\mathsf{\implies x \: = \: \dfrac{-b^{2} \: + \: a^{2} \: \pm \: \sqrt{(b^{2} )^{2} \: + \: (a^{2}) {}^{2} \: + \: 2 \: \cdot \: {a}^{2} \: \cdot {b}^{2} \: }}{2 {a}^{2} {b}^{2} }}$




$\mathsf{\implies x \: = \: \dfrac{-b^{2} \: + \: a^{2} \: \pm \: \sqrt{(b^{2} + \: {a}^{2} ) {}^{2} }}{2 {a}^{2} {b}^{2} }}$




$\mathsf{\implies x \: = \: \dfrac{-b^{2} \: + \: a^{2} \: \pm \: (b^{2} + \: {a}^{2} ) }{2 {a}^{2} {b}^{2} }}$




$\textsf{Taking +ve sign : } \\ \\ \mathsf{\implies x \: = \: \dfrac{ \cancel{-b^{2}} \: + \: a^{2} \: + \: \cancel{ b^{2}} \: + \: a^{2}}{2a^{2}b^{2}}}$




$\mathsf{\implies x \: = \: \dfrac{ \cancel{2a^{2}}}{ \: \: \cancel{2a^{2}}b^{2} \: \: }}$




$\mathsf{\therefore \quad x \: = \: \dfrac{ 1}{ \: b^{2}}}$




$\textsf{Taking - ve sign : } \\ \\ \mathsf{\implies x \: = \: \dfrac{ {-b^{2}} \: + \: \cancel{a^{2}} \: - \: { b^{2}} \: - \: \cancel{a^{2}}}{2a^{2}b^{2}}}$




$\mathsf{\implies x \: = \: \dfrac{ \cancel{ - 2b^{2}}}{ \: \: \cancel{2b^{2}}a^{2} \: \: }}$




$\mathsf{\therefore \quad x \: = \: \dfrac{ - 1}{ \: a^{2}}}$




$\mathsf{\implies x \: = \: \dfrac{-1}{\: a^{2}} \: or \: \dfrac{1}{\: b^{2} \: }}$

Answer 2

Answer :-

x = 1 / a² ; x = 1 / b²



Explaination :-


EQUATION = a² b² x² + b² x - a² x - 1

Solution :-

a² b² x² + b² x - a² x - 1 = 0

b² x ( a² x + 1 ) - 1 ( a² x + 1 ) = 0

( a² x + 1 ) ( b² x - 1 ) = 0

Now , Use Zero Product Rule

a² x + 1 = 0 ; b² x - 1 = 0

a² x = - 1 ; b² x = - 1

x = 1 / a² ; x = 1 / b²