Roots of quadratics.

**grgrsanjay** · August 8th 2010, 05:26 PM

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

**pickslides** · August 8th 2010, 05:33 PM

What are the roots of these equations?

Do you know how to find them?

Hint: $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

**grgrsanjay** · August 8th 2010, 05:35 PM

i know to find them

**eumyang** · August 8th 2010, 05:48 PM

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...

Can you take it from here?

**eumyang** · August 8th 2010, 05:56 PM

(Duplicate post).

Thread: Roots of quadratics.

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