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Find all real values of the parameter $a$ for which the equation $x(x + 1)(x + a)(x + a+ 1) = a^2$ has four real roots.

My Attempt enter image description here enter image description here enter image description here enter image description here

Is my attempt is correct and also is there any other way to solve this problem.

Abhishek Kumar
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    Your approach is great! There are quicker ways to find the critical points $\pm2\pm\sqrt5$, but other than that, I like what you've done here. – Rushabh Mehta Aug 08 '19 at 17:30

1 Answers1

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Here is a simpler solution, obtained by recognizing a certain symmetry.

Let us give names to the LHS and RHS of the given equation :

$$\underbrace{x(x+1)(x+a)(x+a+1)}_{f_a(x)}=\underbrace{a^2}_{g_a(x)} \tag{1}$$

Here are, for two cases ($a=-3$ and $a=-4.5$), the joint graphical representations of the curves of $f_a$ (blue) and $g_a$ (red). The midpoint $x_a:=-\tfrac{a+1}{2}$ of the roots of equation $f_a(x)=0$ is materialized as a little red circle.

enter image description here

Fig. 1 : On the left : case $a=-3$. On the right, case $a=-4.5$. In the first case, the line intersects the curve of $f_a$ only twice, meaning two real roots to equation (1) instead of four. In the second case, there are four intersection points, giving four real roots.

Whatever the value of $a$, the graphical representation of function $f_a$ will always be the same : symmetrical with respect to vertical axis with equation $x=x_a$, with a central maximum with coordinates $(x_a,f(x_a))$. The necessary and sufficient condition for having 4 roots to (1) is clearly that the ordinate of the central maximum is greater than $a^2$, i.e., $f_a(x_a)>a^2$ (see Remark 1 below). An easy calculation gives :

$$\dfrac{(a^2-1)^2}{16} > a^2$$

$$a^4-2a^2+1 > 16 a^2$$

Otherwise said :

$$(a^2-9)^2>80$$

giving :

$$a^2 \notin [9-4\sqrt{5},9+4\sqrt{5}]$$

As $9\pm 4\sqrt{5}=(2 \pm \sqrt{5})^2$, we find back your solution.

Remarks :

  • 1) a fully rigorous approach would be achieved by the following change of variable in (1) : $x=X-\tfrac{a+1}{2}$ giving

$$F(X):=X^4 - \tfrac12(a^2+1)X^2 + \tfrac{1}{16}(a^2-1)^2=(X^2-\tfrac{(a-1)^2}{4})(X^2-\tfrac{(a+1)^2}{4})$$

$F$ is an even "bi-quadratic" function, with variations like this

$$^{+\infty} \ \searrow \ _{_m} \ \nearrow \ ^{F(0)=M} \ \searrow \ _{_m} \ \nearrow \ ^{+\infty}$$

where $M>0$ and $m<0$.

Jean Marie
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