Question

please answer this question fast
with process​

Answer 1

${\huge{\red{\underline{Answer:}}}}$

Option C

$x = ( \frac{b}{a} , - \frac{d}{c} )$

${\huge{\blue{\underline{Given:}}}}$

f(x) = (ax - b)(cx + d) = 0

${\huge{\pink{\underline{To\:Find:}}}}$

The solution of f(x)

${\huge{\orange{\underline{Formula:}}}}$

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{ 2a }$

${\huge{\green{\underline{Solution:}}}}$

(ax - b)(cx + d) = 0

(ac)x² - (bc)x + (ad)x - bd = 0

(ac)x² + (ad - bc)x + (-bd) = 0

Compare with "ax² + bx + c = 0"

$x = \frac{ - b ± \sqrt{ {b}^{2} - 4ac} }{ 2a }$

$x = \frac{ - (ad - bc) ± \sqrt{(ad - bc)2 - 4(ac)( - bd)} }{2(ac) } \\ = \frac{bc - ad ± \sqrt{ {(ad)}^{2} + {(bc)}^{2} - 2abcd + 4abcd } }{2ac} \\ = \frac{bc - ad ± \sqrt{ {(ad)}^{2} + {(bc)}^{2} + 2abcd } }{2ac} \\ = \frac{bc - ad ±( ad + bc)}{2ac}$

1st case:

$x = \frac{bc - ad +( ad + bc)}{2ac} \\ = \frac{2bc}{2ac} = \frac{b}{a} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

2nd case:

$x = \frac{bc - ad -( ad + bc)}{2ac} \\ = - \frac{2ad}{2ac} = - \frac{d}{c} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

${\red{Answer: \: x = ( \frac{b}{a}, - \frac{d}{c} ) }}$

Answer 2

Option C

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