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| 1. |
X2-2x-8 find the relationship between the zeroes and coefficient. |
| Answer» Let p(x) = x2 - 2x - 8By the method of splitting the middle term,{tex}x ^ { 2 } - 2 x - 8 = x ^ { 2 } - 4 x + 2 x - 8{/tex}{tex}= x ( x - 4 ) + 2 ( x - 4 ) = ( x - 4 ) ( x + 2 ){/tex}For zeroes of p(x),p(x) = 0{tex}\\Rightarrow ( x - 4 ) ( x + 2 ) = 0{/tex}{tex}\\Rightarrow x - 4 = 0 \\text { or } x + 2 = 0{/tex}{tex}\\Rightarrow x = 4 \\text { or } x = - 2{/tex}{tex}\\Rightarrow x = 4 , - 2{/tex}So, the zeroes of p(x) are 4 and -2.We observe that, Sum of its zeroes= 4 + (-2) = 2{tex}= \\frac { - ( - 2 ) } { 1 } = \\frac { - \\text { (Coefficient of } x ) } { \\text { Coefficient of } x ^ { 2 } }{/tex}Product of its zeroes{tex}= 4 x ( - 2 ) = - 8 = \\frac { - 8 } { 1 } = \\frac { \\text { Constant term } } { \\text { Coefficient of } x ^ { 2 } }{/tex}Hence,\xa0relation\xa0between zeroes and coefficients is verified. | |