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Solution:-let $\alpha,\beta,\gamma,\delta$ be the roots of equation. It is given that,$\alpha+\beta=0$ and,$\alpha+\beta+\gamma+\delta=3/6=1/2$,this implies $\gamma+\delta=1/2$

and $(\alpha+\beta)(\gamma+\delta)+\alpha\beta+\gamma\delta=8/6=4/3$,this implies $\alpha\beta+\gamma\delta=8/6=4/3$ From here how to proceed,please guide me.

Any suggestion is greatly appreciated.

Styles
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5 Answers5

5

If $\pm \alpha$ are roots, average the statements you know to get

$$6 \alpha^4 + 8 \alpha^2 + 2 = 0$$

which is quadratic in $\alpha^2$, and can be solved for $\alpha$. Now given $\pm \alpha$, the problem can be reduced to another quadratic.

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By Vieta's formulas, $$\alpha+\beta+\gamma+\delta=\frac 12\tag1$$ $$\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=\frac 16\tag2$$ $$\alpha\beta\gamma\delta=\frac 13\tag3$$

From $(1)$, if $\alpha+\beta=0$, then we have $\beta=-\alpha$ and $\delta=1/2-\gamma$.

So, from $(2)$, $$\frac 16=\alpha\beta(\gamma+\delta)+(\alpha+\beta)\gamma\delta=-\frac{\alpha^2}{2}$$ to have $$\alpha^2=-\frac 13\quad\Rightarrow\quad \alpha=\pm\frac{i}{\sqrt 3}$$

Also, from $(3)$, $$\frac 13=\alpha\beta\gamma\delta=\frac 13\gamma\left(\frac 12-\gamma\right)$$ to have $$2\gamma^2-\gamma+2=0\quad\Rightarrow\quad \gamma=\frac{1\pm\sqrt{15}\ i}{4}$$

Therefore, the four roots are $$\pm\frac{i}{\sqrt 3},\ \frac{1\pm\sqrt{15}\ i}{4}$$

mathlove
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3

The given equation is $6x^4 - 3x^3 + 8x^2 -x + 2 = 0$

Given: $\alpha + \beta = 0$ Hence, $\alpha = -\beta$

Now, from the given equation

$\alpha + \beta + \gamma + \delta = 3/6 = 1/2$

$(\alpha + \beta)(\gamma + \delta) + \alpha \beta + \gamma \delta = 8/6 = 4/3$

$(\alpha + \beta) \gamma \delta + \alpha \beta (\gamma + \delta) = 1/6 $

$\alpha \beta \gamma \delta = 2/6 = 1/3$

Applying the given condition, the above equations become,

$\gamma + \delta = 1/2$ ---(I)

$\alpha \beta + \gamma \delta = 4/3$ ---(II)

$\alpha \beta (\gamma + \delta) = 1/6 $ ---(III)

$\alpha \beta \gamma \delta = 1/3$ ---(IV)

using I and III, we get, $\alpha \beta = 1/3$ ---(V)

using IV and V, we get, $\gamma \delta = 1$ ---(VI)

So, now we have 2 quadratic equations,

$x^2 - 0.x + 1/3 = 0$ (for $\alpha , \beta$) and $x^2 - 1/2.x + 1 = 0$ (for $\gamma , \delta$)

ie: $3x^2 + 1 = 0$ and $2x^2 - x + 2 = 0$

Solving, the quadratic equations, we get, the roots as

$$\frac {i}{\sqrt 3}, -\frac {i}{\sqrt 3}, \frac{1 + i\sqrt 15}{4}, \frac{1 - i\sqrt 15}{4}$$

Shikhar S. Maheshwari
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The polynomial easily factors as $(3x^2+1)(2x^2-x+2)$. We can solve each quadratic in the usual way to get $x=\pm\frac{i}{\sqrt3},\frac{1}{4}\pm\frac{i\sqrt{15}}{4}$.

almagest
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  • Doesn't answer the question of how to use the fact that two roots add to zero to solve. Knowing two roots add to zero means you don't have to guess how to "easily factor". – fleablood May 31 '16 at 06:11
  • That was not part of the question - if you want to be pedantic, it was even weaker than the classic "hence or otherwise". But the fact that it has roots $\pm k$ means that $x^2-k^2$ is a factor, which helps you to factor, if you need help. – almagest May 31 '16 at 06:15
  • "that's not part of the question" ????? That's the entirety of the question! – fleablood May 31 '16 at 15:46
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In my opinion, the most straightforward way to solve problems like this (and that) is with polynomial GCDs. These can be computed with the euclidean algorithm, without need for searching or guessing.

Set $p(x)=6x^4-3x^3+8x^2-x+2$. If $x=\pm\alpha$ is a solution of $p(x)=0$, then the polynomials $p(x)$ and $q(x)=p(-x)$ have a common zero, therefore $p$ and $q$ have a nontrivial polynomial GCD. Here, it turns out that that GCD (up to a constant nonzero factor) is $$g(x)=3x^2+1,$$ the roots of which give precisely the $\pm\alpha$ sought.

ccorn
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