# Quadratic equation with roots of opposite signs

• brotherbobby
In summary, the equation 2x^2-(a^3+8a-1)x+a^2-4a = 0 has roots of opposite signs. When solving for a, the constant term must be less than 0 in order for the roots to be of opposite signs. If a is greater than 4, then the roots will not be of opposite signs and there is no answer provided in the book.
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.

brotherbobby said:
What is the value of a ?
It asks for a range, not a value!
brotherbobby said:
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 ?

docnet and Delta2
You should look for values of ##a## such that the product of the roots becomes negative.

docnet
brotherbobby said:
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.

Last edited:
docnet
brotherbobby said:
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##?

Delta2
BvU said:
What is a sufficient condition that two real numbers α,β are of opposite sign ?

That their product is negative?

BvU
docnet said:
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.

docnet and PeroK

## 1. What is a quadratic equation with roots of opposite signs?

A quadratic equation with roots of opposite signs is a polynomial equation of degree 2 with two solutions that have opposite signs when solved. It can be written in the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable.

## 2. How do you solve a quadratic equation with roots of opposite signs?

To solve a quadratic equation with roots of opposite signs, you can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. First, identify the values of a, b, and c from the equation. Then, substitute these values into the formula and solve for x. The two solutions will have opposite signs.

## 3. What does it mean when a quadratic equation has roots of opposite signs?

When a quadratic equation has roots of opposite signs, it means that the solutions to the equation are two real numbers that have opposite signs. This indicates that the parabola represented by the equation crosses the x-axis at two different points, one on each side of the y-axis.

## 4. Can a quadratic equation have roots of opposite signs and still have a positive discriminant?

Yes, a quadratic equation can have roots of opposite signs and still have a positive discriminant. The discriminant, b^2 - 4ac, determines the nature of the solutions. If the discriminant is positive, the equation will have two distinct real solutions, which can have opposite signs.

## 5. What is the significance of a quadratic equation with roots of opposite signs?

A quadratic equation with roots of opposite signs has significance in graphing and solving problems in physics, engineering, and other fields. The solutions to the equation represent the x-intercepts of the parabola, and the opposite signs indicate that the parabola crosses the x-axis at two different points. This information can be used to analyze the behavior of a system or to find the maximum or minimum value of a function.

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