Question

solve the equation .........

Answer 1

x - b - x by x^2 - xb = a - b - a by a^2 - ab

- b by x ^ 2 - xb = - b by a^ 2 - ab

- b ( a^2 - ab ) = - b ( x^2 - xb )

a^2 - ab = x^2 - xb

a ( a - b ) = x ( x - b )

sorry bro but isk baad aap solve krlo.....

Answer 2

$\underline\bold{\huge{ANSWER \: :}}$


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▶The general form of the quadratic equation is (ax²+bx+c) = 0, for any problems, but (a ≠ 0)

▶ Every quadratic equation has two and only two roots.

▶The term "b²-4ac" is called discriminant of the quadratic equation.

▶ When b²-4ac = 0, then the roots are real & equal.

▶ When b²-4ac > 0, then the roots are real and distinct.

▶ When b²-4ac < 0, then there will be no real roots.

▶ Sum of the roots of a quadratic equation is always -b/a.

▶ Product of the roots of a quadratic equation is always c/a.

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✔✔✔✔✔ SOLUTION ✔✔✔✔✔

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$\frac{1}{x} - \frac{1}{x - b} = \frac{1}{a} - \frac{1}{a - b}$


=> $\frac{x - b - x}{x(x - b)} = \frac{a - b - a}{a(a - b)}$


=> $\frac{ - b}{x(x - b)} = \frac{ - b}{a(a - b)}$


=> $\frac{1}{ {x}^{2} - \: bx } = \frac{1}{ {a}^{2} - \: ab }$


=> a² - ab = x² - bx


=> a² - x² - ab + bx = 0


=> ( a + x ) ( a - x ) - b ( a - x ) = 0


=> ( a - x ) ( a + x - b ) = 0

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$\bold{\huge{ULTIMATELY}}$

$\bold{EITHER}$

a + x - b = 0

=> x = (b - a)

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$\bold{OR}$

a - x = 0

=> a = x