Quadratic equation with roots of opposite signs

[brotherbobby]

Homework Statement

At what values of $a$ does the equation $2x^2-(a^3+8a-1)x+a^2-4a = 0$ possess roots of opposite signs?

Relevant Equations

For a quadratic equation $ax^2+bx+c=0$ having roots $\alpha,\beta$, the sum of the roots $\alpha+\beta = -\frac{b}{a}$ and product of the roots $\alpha\beta = \frac{c}{a}$.

Given : The equation $2x^2-(a^3+8a-1)x+a^2-4a = 0$ with roots of opposite signs.

Required : What is the value of $a$ ?

Attempt : The roots of the equation must be of the form $\alpha, -\alpha$. The sum of the roots $0 = a^3+8a-1$.

I do not know how to solve this equation.

However, on plotting the graph of this function [$f(x) = x^3+8x-1$], I find that $a = 0.125$.

However, this is not the answer in the book.

Answer : $a \in (0;4)$ (from book)

Any help would be welcome.

 

[BvU]

What is the value of a ?

It asks for a range, not a value!

The roots of the equation must be of the form $\alpha, -\alpha$ The sum of the roots $0=a^3+8a−1$.

You make that up ! Nowhere is said that the absolute values should be equal !

What is a sufficient condition that two real numbers $\alpha, \beta$ are of opposite sign ?

 

[Delta2]

You should look for values of $a$ such that the product of the roots becomes negative.

 

[PeroK]

Homework Statement:: At what values of $a$ does the equation $2x^2-(a^3+8a-1)x+a^2-4a = 0$ possesses roots of opposite signs?
Relevant Equations:: For a quadratic equation $ax^2+bx+c=0$ having roots $\alpha,\beta$, the sum of the roots $\alpha+\beta = -\frac{b}{a}$ and product of the roots $\alpha\beta = \frac{c}{a}$.

You have $a$ already in the original equation, so you shouldn't really use $a$ for the first coefficient here as well. You can and should simply write $2x^2 + bx + c$ here.

Hint: perhaps focus on $c$ and the quadratic forumula.

 

[docnet]

Given : The equation $2x^2-(a^3+8a-1)x+a^2-4a = 0$ with roots of opposite signs.

Required : What is the value of $a$ ?

Written in standard form the quadratic equation becomes

$x^2-\frac{(a^3+8a-1)}{2}x+\frac{a^2-4a}{2}=0$

The rule is, if the constant term is $<0$ then the roots have opposite signs.

So for which values of a does the constant term $\frac{a^2-4a}{2}<0$?

 

[brotherbobby]

What is a sufficient condition that two real numbers α,β are of opposite sign ?

That their product is negative?

 

[brotherbobby]

Written in standard form the quadratic equation becomes

$x^2-\frac{(a^3+8a-1)}{2}x+\frac{a^2-4a}{2}=0$

The rule is, if the constant term is $<0$ then the roots have opposite signs.

So for which values of a does the constant term $\frac{a^2-4a}{2}<0$?

Thank you for your help. I have got it. Let me do the problem still for completeness' sake.

We have the equation $2x^2-(a^3+8a-1)x+a^2-4a = 0$ possessing roots of opposite signs.

We are asked for what values of $a$ will they do so?

Let the roots be $\alpha, \beta$. They are of opposite signs, hence $\alpha \beta = -\text{ve}$.

Original equation can be simplified to : $x^2 - \frac{a^3+8a-1}{2}x + \frac{a^2-4a}{2} = 0$.

Product of the roots $\alpha \beta = \frac{a^2-4a}{2} = -\text{ve}$.

Thus we have to find the value of $a$ for which $a^2-4a = -\text{ve}$ (since 2 times a negative number is negative.)

Continuing from above : $a(a-4) = -\text{ve} \Rightarrow 0\le a \le 4\Rightarrow \boxed{a \in (0;4)}$, matching the answer from the book.

Thank you all very much.