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| 1. |
Solve the following quadratic equation by factorization method:9x*x-6b*bx-(a*a*a*a-b*b*b*b) |
| Answer» We have,9x2 - 6b2x - (a4 - b4) = 0Here, Constant term = a4 - b4 = (a2 - b2)(a2\xa0+ b2)Also, the coefficient of middle term is - 6b2 = [3(a2 + b2) - 3(a2\xa0- b2)]Now using the above two values in given equation, 9x2 - 6b2x - (a4 - b4) = 0{tex}{/tex}\xa0we have, 9x2 - [3(a2 + b2) - 3(a2 - b2)]x - (a2 - b2)(a2\xa0+ b2) = 0{tex}\\Rightarrow{/tex} 9x2 - 3(a2\xa0+ b2)x + 3(a2 - b2)x - (a2 - b2)(a2 + b2) = 0{tex}\\Rightarrow{/tex} 3x[3x - (a2\xa0+ b2)] + (a2 - b2)[3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex} [3x + (a2\xa0- b2)][3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex}either [3x + (a2\xa0 - b2)] = 0 or, [3x - (a2 + b2)] = 0{tex}\\Rightarrow{/tex} 3x = -(a2 - b2) or 3x = a2 + b2{tex}\\Rightarrow x = -(\\frac{{{a^2} - {b^2}}}{3}){/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}{tex}\\Rightarrow x = \\frac{{{b^2} - {a^2}}}{3}{/tex} or {tex}x = \\frac{{{a^2} + {b^2}}}{3}{/tex}Hence, the roots of given quadratic equation are\xa0{tex}\\frac{{b^2 - a^2}}{3}{/tex}and\xa0{tex}\\frac{{a^2 + b^2}}{3}{/tex} | |