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| 1. |
X2-(2b-1)x+(b2-b-20)=0 find the value of x |
| Answer» In the given quadratic equation,\xa0x2 - (2b - 1 ) x + (b2- b - 20) = 0A= +1, B=\xa0- (2b - 1 ), C=\xa0(b2- b - 20){tex}x = {-B \\pm \\sqrt{B^2-4AC} \\over 2A}{/tex}{tex}x = \\frac { ( 2 b - 1 ) \\pm \\sqrt { ( 2 b - 1 ) ^ { 2 } - 4 \\left( b ^ { 2 } - b - 20 \\right) } } { 2 }{/tex}{tex}x = \\frac { ( 2 b - 1 ) \\pm 9 } { 2 }{/tex}{tex}x = \\frac { 2 b + 8 } { 2 } \\quad and \\quad x = \\frac { 2 b - 10 } { 2 }{/tex}{tex}\\therefore x= b+4 \\quad and \\quad x=b-5{/tex}\xa0 | |