Saved Bookmarks
| 1. |
Solved by the complete square method a²x²-3abx+2b²=0 |
| Answer» The given equation is:a2x2\xa0- 3abx + 2b2 = 0{tex}\\Rightarrow{/tex}\xa0a2x2\xa0- 3abx = -2b2Adding\xa0{tex}\\left( \\frac { 3 b } { 2 } \\right) ^ { 2 }{/tex} on both sides, we have,{tex}(ax)^2-2\\times(ax)\\times\\frac{3b}2+\\left(\\frac{3b}2\\right)^2=-2b^2+\\left(\\frac{3b}2\\right)^2{/tex}{tex}(ax-\\frac{3b}2)^2=-2b^2+(\\frac{3b}2)^2{/tex}{tex}(ax-\\frac{3b}2)^2=\\frac{b^2}4{/tex}\xa0{tex}\\Rightarrow \\quad \\left( a x - \\frac { 3 b } { 2 } \\right) = \\pm \\frac { b } { 2 }{/tex}\xa0(By taking square root on both sides)So either,\xa0{tex} \\left( a x - \\frac { 3 b } { 2 } \\right) = \\frac { b } { 2 }{/tex}\xa0or\xa0{tex}\\left( a x - \\frac { 3 b } { 2 } \\right) = \\frac { - b } { 2 }{/tex}{tex}\\Rightarrow \\quad a x = \\left( \\frac { b } { 2 } + \\frac { 3 b } { 2 } \\right) = \\frac { 4 b } { 2 }{/tex}= 2b,thus,\xa0x = 2b/aor ax =\xa0{tex}\\left( \\frac { - b } { 2 } + \\frac { 3 b } { 2 } \\right) = \\frac { 2 b } { 2 }{/tex}\xa0= bthus, x = b/aSo,\xa0x =\xa0{tex}\\frac { 2 b } { a }{/tex}\xa0or x =\xa0{tex}\\frac { b } { a }{/tex}.Hence,\xa0the roots of the given equation are\xa0{tex}\\frac { 2 b } { a }{/tex}\xa0and\xa0{tex}\\frac { b } { a }{/tex}. | |