Question

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$ab {x}^{2} + ( {b}^{2} - ac)x - bc = 0$
${solve \: this \: by \: quadratic \: formula \: }$
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❤️Have a wonderful day ❤️

Answer 1

Hope it will help u...

Answer 2

Hey Friends!!

Here is your answer↓⬇


⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇⬇


▶The given equation is:-)

$\bf{ab {x}^{2} + ( {b}^{2} - ac)x - bc = 0.}$

▶This is of the form Ax² + Bx + C = 0, where

↪➡ A = ab,

↪➡ B = ( b² - ac ) and

↪➡ C = -bc.


▶Now,

=> D = ( B² - 4AC ).

=> D = ( b² - ac )² + 4ab²c.

=> D = b⁴ + a²c² - 2ab²c + 4ab²c.

=> D = b⁴ + a²c² + 2ab²c.

=> D = ( b² + ac)² > 0.

So, the given equation has real roots.

▶ Now,


$\huge \bf = > \sqrt{ D } = ( {b}^{2} + ac).$


▶ By using quadratic formula:-)

$\huge \bf \: \alpha = \frac{ - B + \sqrt{ D} }{2 A }$



$\bf = > \alpha = \frac{ - ( {b}^{2} - ac) + ( {b}^{2} + ac) }{2ab} .$

$\bf = > \alpha = \frac{2ac}{2ab} .$

$\huge \boxed{= > \alpha = \frac{c}{b} .}$


▶ By using 2nd quadratic formula:-)


$\huge \bf \: \beta = \frac{ - B - \sqrt{ D } }{2 A } .$


$\bf = > \beta = \frac{ - ( {b}^{2} - ac) + ( {b}^{2} + ac) }{2ab} .$


$\bf = > \beta = \frac{ - 2 {b}^{2} }{2ab} .$


$\huge \boxed{ = > \beta = \frac{ - b}{a}. }$



✅✅Hence, c/b and -b/a are the roots of the given equation✔✔.




$\huge \boxed{THANKS}$



$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: y ou}$