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Find all real values of a for which the equation \((x^2 + a)^2 + a = x\) has four real roots.

 Feb 21, 2019

Best Answer 

 #2
avatar+22188 
+4

Find all real values of a for which the equation \(\mathbf{(x^2 + a)^2 + a = x}\) has four real roots.

 

My attempt:

I assume there are 4 distinct real roots. So there are 3 local maxima/minima.

We find the local maxima/minima by differentiation. Maxima/minima occur when f'(x) = 0

 

1. Differentiation:

\(\begin{array}{|rcll|} \hline y &=& (x^2 + a)^2 -x+a \\ y' &=& 2(x^2+a)\cdot 2x-1 \\ 0 &=& 4x(x^2+a) - 1 \\ 4x(x^2+a) &=& 1 \quad | \quad : 4 \\ x(x^2+a) &=& \frac14 \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

 

The zero set of discriminant of the cubic  \(\mathbf{Ax^{3}+Bx^{2}+Cx+D\,}\) has discriminant
  \(\mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

 

\(\begin{array}{|rcll|} \hline \mathbf{Ax^{3}+Bx^{2}+Cx+D\,} \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \quad | \quad A=1,\ B=0,\ C=a,\ D=-\frac14 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD } &>& 0 \quad | \quad \text{there are three distinct real numbers} \\ -4a^3 -27\left(-\frac14\right)^2 &>& 0 \\ -4a^3 -\frac{27}{16} &>& 0 \\ -4a^3 &>& \frac{27}{16} \quad | \quad : -4 \\ a^3 &<& -\frac{27}{64} \\ a^3 &<& -\frac{3^3}{4^3} \\ \mathbf{ a } &\mathbf{<}& \mathbf{-\frac{3}{4}} \\ \hline \end{array}\)

 

\(\mathbf{(x^2 + a)^2 + a = x}\) has four distinct real roots if  \(\mathbf{ a<-\frac{3}{4}}\)

 

laugh

 Feb 22, 2019
edited by heureka  Feb 22, 2019
 #1
avatar+194 
0

x^5, a^3

 

 

Maybe right 

 

Not sure about this 

 Feb 21, 2019
 #2
avatar+22188 
+4
Best Answer

Find all real values of a for which the equation \(\mathbf{(x^2 + a)^2 + a = x}\) has four real roots.

 

My attempt:

I assume there are 4 distinct real roots. So there are 3 local maxima/minima.

We find the local maxima/minima by differentiation. Maxima/minima occur when f'(x) = 0

 

1. Differentiation:

\(\begin{array}{|rcll|} \hline y &=& (x^2 + a)^2 -x+a \\ y' &=& 2(x^2+a)\cdot 2x-1 \\ 0 &=& 4x(x^2+a) - 1 \\ 4x(x^2+a) &=& 1 \quad | \quad : 4 \\ x(x^2+a) &=& \frac14 \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \\ \hline \end{array}\)

 

The zero set of discriminant of the cubic  \(\mathbf{Ax^{3}+Bx^{2}+Cx+D\,}\) has discriminant
  \(\mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD\,.}\)
The discriminant is zero if and only if at least two roots are equal.
If the coefficients are real numbers, and the discriminant is not zero,
the discriminant is positive if the roots are three distinct real numbers,
and negative if there is one real root and two complex conjugate roots.

 

\(\begin{array}{|rcll|} \hline \mathbf{Ax^{3}+Bx^{2}+Cx+D\,} \\ \mathbf{x^3+ax-\frac14} & \mathbf{=} & \mathbf{0} \quad | \quad A=1,\ B=0,\ C=a,\ D=-\frac14 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ B^{2}C^{2}-4AC^{3}-4B^{3}D-27A^{2}D^{2}+18ABCD } &>& 0 \quad | \quad \text{there are three distinct real numbers} \\ -4a^3 -27\left(-\frac14\right)^2 &>& 0 \\ -4a^3 -\frac{27}{16} &>& 0 \\ -4a^3 &>& \frac{27}{16} \quad | \quad : -4 \\ a^3 &<& -\frac{27}{64} \\ a^3 &<& -\frac{3^3}{4^3} \\ \mathbf{ a } &\mathbf{<}& \mathbf{-\frac{3}{4}} \\ \hline \end{array}\)

 

\(\mathbf{(x^2 + a)^2 + a = x}\) has four distinct real roots if  \(\mathbf{ a<-\frac{3}{4}}\)

 

laugh

heureka Feb 22, 2019
edited by heureka  Feb 22, 2019
 #3
avatar+100571 
+2

Thanks for that one, Heureka....a lot of good info here ....definitely one to put on my "Watchlist" to review!!!

 

 

cool cool cool

CPhill  Feb 22, 2019
 #4
avatar+22188 
+3

Thank you, CPhill !!!

 

laugh

heureka  Feb 25, 2019

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