InterviewSolution
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InterviewSolution
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Solve the equation `(x+2)(x+3)(x+8)xx(x+12)=4x^2dot` |
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Answer» Since `(-2)(-12)=(-3)(-8)` so we can we write given equation as `(x+2)(x+12)(x+3)(x+8)=4x^(2)` `implies(x^(2)+14x+24)(x^(2)+11x+24)=4x^(2)`……I Now `x=0` is not a root of given equation on dividing by `x^(2)` in both sides of Eq. (i) we get `(x+24/x+14)(x+24/x+11)=4` ........ii Put `x+24/x=y` then Eq. (ii) can be reduced in the form `(y+14)(y+11)=4` or `y^(2)+25y+150=0` `:.y_(1)=-15` and `y_(2)=-10` Thus, the original equation is equivalent to the collection of equatiosn `[(x+24/x=-15),(x+24/x=-10):]` i.e. `[(x^(2)+15x+24=0),(x^(2)+10x+24=0):]` On solving these collection we get `x_(1)=(-15-sqrt(129))/2,x_(2)=(-15+sqrt(129))/2,x_(3)=-6,x_(4)=-4` |
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