A) \[\frac{b}{ac}\] and \[\frac{c}{b}\]
B) \[\frac{ab}{c}\] and \[\frac{a}{b}\]
C) \[\frac{-b}{a}\]and \[\frac{c}{b}\]
D) \[\frac{b}{a}\] and \[\frac{-c}{b}\]
Correct Answer: C
Solution :
| \[ab{{x}^{2}}+\left( {{b}^{2}}-ac \right)x-bc=0\] |
| \[\Rightarrow \,ab{{x}^{2}}-acx+{{b}^{2}}x-bc=0\] |
| \[\Rightarrow \,ax\left( bx-c \right)+b\left( bx-c \right)=0\] |
| \[\Rightarrow \,\,\left( ax+b \right)\left( bx-c \right)=0\] |
| Hence, \[x=-\frac{b}{a},\,\frac{c}{b}\] are zeroes. |
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