Question

solve for x.
${a}^{2} {b}^{2} {x}^{2} + {b}^{2}x - {a}^{2} x - 1 = 0$
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Answer 1

$\huge\text{\underline{Answer}}$

$\bold{x = - \frac{1}{ {a}^{2} } }$

$\bold{x = \frac{1}{ {b}^{2} } }$

$\sf{\underline{Step by step explanation}}$

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

We have to find the value of X.

it is a quadratic polynomial so ,it has two zeroes.

Now take common,

$\implies$ $\bold{{b}^{2} x( {a}^{2} x + 1) - 1( {a}^{2}x + 1) = 0 }$

$\implies$ $\bold{ ( {a}^{2} x + 1)( {b}^{2} x - 1) = 0 }$

Either,

$\implies$ $\bold{( {a}^{2} x + 1) = 0 \: \: or \: ( {b}^{2} x - 1) = 0}$

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

$\implies$ $\bold{x = \frac{ - 1}{ {a}^{2} } \: \: \: or \: \: x = \frac{1}{ {b}^{2} } }$

hence, $\boxed{\sf{ x = \frac{ - 1}{ {a}^{2} } }}$and $\boxed{\sf{ x = \frac{1}{ {b}^{2} } }}$ are required solutions.

Answer 2

Answer:

The above written answer is correct