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13 Feb 2012 20:48:52 IST

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quadratic!!! please help!!!

Mathematics

for what real values of a do the roots of the equation x^2 -2x - (a^2 - 1) =0 lie between x^2 -2 (a+1)x +a(a-1) =0 ??

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jagdish singh

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14 Feb 2012 20:10:55 IST

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$\hspace{-16}$Here Quadratic Equation is $\mathbf{x^2-2x-(a^2-1)=0}$\\\\ $\mathbf{x^2-2x+1=a^2\Leftrightarrow (x-1)^2=a^2}$\\\\ So Roots are $\mathbf{x-1=\pm a\Leftrightarrow \boxed{\bold{x=1\pm a\forall a\in \mathbb{R}}}}$\\\\ Now Thses Roots are lie b/w $\mathbf{f(x)=x^2-2(a+1)x+a(a-1)=0}$\\\\ Means This is an Upward Parabola and $\mathbf{x=1\pm a}$ lie two roots of $\mathbf{f(x)=0}$\\\\ So Discriminant $\mathbf{\geq 0}$ and $\mathbf{1.f(1\pm a)< 0}$\\\\ Where $\mathbf{1}$ is a Leading Coefficient of $\mathbf{f(x)=0}$ i.e Coeffi. of $\mathbf{x^2}$\\\\ So $\mathbf{D\geq 0\Leftrightarrow (a+1)^2-4.a.(a-1)\leq 0}$\\\\ $\mathbf{3a+1\geq 0\Leftrightarrow \boxed{\bold{a\geq -\frac{1}{3}}}}$\\\\ Similarly $\mathbf{f(1+a)<0}$\\\\ $\mathbf{(1+a)^2-2.(a+1).(a+1)+a(a-1)<0}$\\\\ $\mathbf{3a+1>0\Leftrightarrow \boxed{\bold{a>-\frac{1}{3}}}}$\\\\ Similarly $\mathbf{f(1-a)<0}$\\\\ $\mathbf{(1-a)^2-2.(a+1)(a-1)+a(a-1)<0}$\\\\ $\mathbf{a-1>0\Leftrightarrow \boxed{\bold{a>1}}}$\\\\$

$\hspace{-16}$So from these Condition We Get Common value of $\mathbf{a}$ is $\mathbf{\left(1,\infty\right)}$\\\\ So $\boxed{\boxed{\mathbf{a\in \left(1,\infty\right)}}}$$

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