Condition for A Quartic Equation to have a Real Root

In summary, a quartic equation is a polynomial equation of the fourth degree with the highest exponent being 4. A real root of a quartic equation is a solution that is a real number, not a complex number. To determine if a quartic equation has a real root, the discriminant must be greater than or equal to 0. If a quartic equation does not have a real root, all of its solutions will be complex numbers. Other conditions for a quartic equation to have a real root include having all real coefficients or two pairs of complex conjugate roots.
  • #1
anemone
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Show $20a^2+20b^2+5c^2\ge 64$ if $y=x^4+ax^3+bx^2+cx+4$ has a real root.
 
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Solution of other:
Note that $x=0$ is not a solution for $x^4+ax^3+bx^2+cx+4=0$.

$ax^3+bx^2+cx=-(x^4+4)\\(ax^3+bx^2+cx)^2=-(x^4+4)^2\\\left(2a\dfrac{x^3}{2}+2b\dfrac{x^2}{2}+cx \right)=-(x^4+4)^2 \le (4a^2+4b^2+c^2)\left(\dfrac{x^6}{4}+\dfrac{x^4}{4}+x^2\right)$
by the CauchySchwarz inequality

This gives
$4a^2+4b^2+c^2\ge \dfrac{4(x^4+4)^2}{x^6+x^4+4x^2}$

Let $t=x^2$, now, we have to prove $\dfrac{4(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{64}{5}$, i.e. $\dfrac{(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{16}{5}$.

This is true since $\dfrac{(t^2+4)^2}{t^3+t^2+4t}\ge \dfrac{16}{5}$ implies $5(t^4+8t^2+16)\ge 16t^3+16t^2+64t$, or $(t-2)^2(5t^2+4t+20)\ge 0$.
 

FAQ: Condition for A Quartic Equation to have a Real Root

1. What is a quartic equation?

A quartic equation is a polynomial equation of degree 4, meaning that the highest exponent of the variable in the equation is 4. It is written in the form ax4 + bx3 + cx2 + dx + e = 0, where a, b, c, d, and e are constants and x is the variable.

2. What does it mean for a quartic equation to have a real root?

A real root of a quartic equation is a value of x that makes the equation equal to 0 when substituted in. It is a solution that lies on the real number line, as opposed to imaginary roots which lie on the complex number plane.

3. What is the condition for a quartic equation to have a real root?

The condition for a quartic equation to have a real root is that the discriminant of the equation, b2 - 4ac, must be greater than or equal to 0. This ensures that the equation has at least one real solution.

4. How can I determine if a quartic equation has a real root?

You can determine if a quartic equation has a real root by calculating the discriminant and checking if it is greater than or equal to 0. If it is, then the equation has at least one real solution. You can also graph the equation and see if it intersects the x-axis at any point, which would indicate a real root.

5. Can a quartic equation have more than one real root?

Yes, a quartic equation can have more than one real root. In fact, it can have up to four real roots, as the degree of the equation is 4. However, it is also possible for a quartic equation to have less than four real roots or no real roots at all.

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