Question

Solve by quadratic formula or completion square

13. a^2b^2x^2 - a^2 x - b^2 x + 1 = 0

Ans: 1 / a^2 , 1 / b^2

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Answer 1

☆ Refer to the attachment for solution.

☆ It has been solved by both methods.

Answer 2

Quadratic Formula Method


Let us see the general case. Consider a quadratic equation:

$ax^2+bx+c=0$
The Discriminant is given by:

$D = b^2-4ac$

And the solutions are:

$x = \frac{-b\pm \sqrt{D}}{2a}$

Looking at the question:

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

The Discriminant is:

$D = (-(a^2+b^2))^2-4(a^2b^2)(1) \\ \\ \implies D = (a^2+b^2)^2-4a^2b^2 \\ \\ \implies D = a^4+2a^2b^2+b^4-4a^2b^2 \\ \\ \implies D = a^4-2a^2b^2+b^4 \\ \\ \implies D = (a^2-b^2)^2 \\ \\ \implies \sqrt{D} = a^2-b^2$

Now the solutions are:

$x = \frac{-b\pm\sqrt{D}}{2a} \\ \\ \\ \implies x = \frac{-(-(a^2+b^2)) + (a^2-b^2)}{2a^2b^2} \qquad OR \qquad x = \frac{-(-(a^2+b^2)) - (a^2-b^2)}{2a^2b^2} \\ \\ \\ \implies x = \frac{a^2+b^2+a^2-b^2}{2a^2b^2} \qquad OR \qquad x = \frac{a^2+b^2-a^2+b^2}{2a^2b^2} \\ \\ \\ \implies x = \frac{2a^2}{2a^2b^2} \qquad OR \qquad x = \frac{2b^2}{2a^2b^2} \\ \\ \\ \implies \boxed{x = \frac{1}{b^2}} \qquad OR \qquad \boxed{x = \frac{1}{a^2}}$


Thus, The Solutions are $\bold{\frac{1}{a^2}}$ and $\bold{\frac{1}{b^2}}$