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Solve for quadratic equation by factorisation x-a +x-b = a+b. x-b. X-b. b. a |
| Answer» We have,{tex}\\frac{{x - a}}{{x - b}} + \\frac{{x - b}}{{x - a}} = \\frac{a}{b} + \\frac{b}{a}{/tex}{tex}\\Rightarrow \\frac{{(x - a)(x - a) + (x - b)(x - b)}}{{(x - b)(x - a)}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex} \\Rightarrow \\frac{{{x^2} + {a^2} - 2ax + {x^2} + {b^2} - 2bx}}{{{x^2} - bx - ax + ab}}{/tex}{tex} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow \\frac{{2{x^2} - 2ax - 2bx + {a^2} + {b^2}}}{{{x^2} - bx - ax + ab}} = \\frac{{{a^2} + {b^2}}}{{ab}}{/tex}{tex}\\Rightarrow{/tex}{tex} (2x^2 - 2ax - 2bx + a^2 + b^2)ab = (a^2 + b^2)(x^2 - bx - ax + ab){/tex}{tex}\\Rightarrow{/tex}{tex}2abx^2 - 2a^2bx - 2ab^2x + a^3b + ab^3 = a^2x^2 - a^2bx -a^3x + a^3b + b^2x^2 - b^3x - ab^2x + ab^3{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}2abx^2 - a^2x^2 - a^2bx - ab^2x + a^3x + b^3x - b^2x^2 = 0{/tex}{tex}\\Rightarrow{/tex}{tex}(2ab - a^2 - b^2)x^2 + (-a^2b - ab^2 + a^3 + b^3)x = 0{/tex}{tex}\\Rightarrow{/tex} x[(2ab - a2 - b2)x + (a3 + b3 - a2b - ab2)] = 0{tex}\\Rightarrow{/tex} x = 0 or (2ab - a2 - b2)x + a3 + b3 -a2b - ab2 = 0Now,(2ab - a2 - b2)x + a3 + b3 - a2b - ab2 = 0{tex}\\Rightarrow{/tex} (2ab - a2 - b2)x = a2b + ab2 - a3 - b3{tex}\\Rightarrow{/tex} -(a2 - b2 - 2ab)x = a2b - b3 + ab2 - a3{tex}\\Rightarrow{/tex} -(a - b)2x = b(a2 - b2) + a(b2 - a2){tex}\\Rightarrow{/tex} (a - b)2x = -b(a2 - b2) - a(b2 - a2){tex} \\Rightarrow x = \\frac{{ - b({a^2} - {b^2}) + a({a^2} - {b^2})}}{{{{(a - b)}^2}}}{/tex}{tex}\\Rightarrow x = \\frac{{\\left( {{a^2} - {b^2}} \\right)(a - b)}}{{{{(a - b)}^2}}}{/tex}{tex} \\Rightarrow x = \\frac{{{a^2} - {b^2}}}{{a - b}} = \\frac{{(a - b)(a + b)}}{{(a - b)}}{/tex}{tex}\\Rightarrow{/tex} x = a + b{tex}\\therefore{/tex}\xa0x = 0 or x = a + b | |