• Aug 8th 2010, 05:26 PM
grgrsanjay
if the greater root of the equation $x^2 -(a^2 +a+1)x+a(a^2+1)$=0 lies between the roots of the equation $x^2 -a^2 x-2(a^2 -2)$ =0 then a lies in
(a)[1,2]
(b)[-2,-1]
(c)[-8,-root5]U[root5,I]
(d)none of these
• Aug 8th 2010, 05:33 PM
pickslides
What are the roots of these equations?

Do you know how to find them?

Hint: $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$
• Aug 8th 2010, 05:35 PM
grgrsanjay
i know to find them
• Aug 8th 2010, 05:48 PM
eumyang
The solutions of the quadratic equation
$x^2 -a^2 x-2(a^2 -2) = 0$ are
\begin{aligned}
x &= \dfrac{a^2 \pm \sqrt{a^4 - 4(1)(2)(a^2 - 2)}}{2} \\
&= \dfrac{a^2 \pm \sqrt{a^4 - 8a^2 + 16}}{2} \\
&= \dfrac{a^2 \pm (a^2 - 4)}{2} \\
&= 2,\:a^2 - 2
\end{aligned}

The solutions of the quadratic equation
$x^2 -(a^2 +a+1)x+a(a^2+1) = 0$ are
\begin{aligned}
x &= \dfrac{a^2 + a + 1 \pm \sqrt{(-(a^2 +a + 1))^2 - 4(1)(a)(a^2 + 1)}}{2} \\
&= \dfrac{a^2 + a + 1 \pm \sqrt{a^4 + 2a^3 + 3a^2 + 2a + 1 -4a^3 - 4a}}{2} \\
&= \dfrac{a^2 + a + 1 \pm \sqrt{a^4 - 2a^3 + 3a^2 - 2a + 1}}{2} \\
&= \dfrac{a^2 + a + 1 \pm (a^2 - a + 1)}{2} \\
&= a,\:a^2 + 1
\end{aligned}

The greater root here is a^2 + 1.
...

Jeez, I could have "factored" these... (Headbang)
Can you take it from here?
• Aug 8th 2010, 05:56 PM
eumyang
(Duplicate post).