Find-the-real-zeros-of-the-polynomial-P-a-x-x-2-1-x-1-2-ax-2-where-a-is-a-given-real-number-

Question Number 152544 by liberty last updated on 29/Aug/21

Find the real zeros of the polynomial

P_{a} (x) = (x^{2} + 1) {(x - 1)}^{2} - {ax}^{2}

where a is a given real number

Commented by MJS_new last updated on 30/Aug/21

I get

x_{1} = \frac{1 - \sqrt{a + 1} - \sqrt{a - 2 - 2 \sqrt{a + 1}}}{2} \in R with a ⩾ 8

x_{2} = \frac{1 - \sqrt{a + 1} + \sqrt{a - 2 - 2 \sqrt{a + 1}}}{2} \in R with a ⩾ 8

x_{3} = \frac{1 + \sqrt{a + 1} - \sqrt{a - 2 + 2 \sqrt{a + 1}}}{2} \in R with a ⩾ 0

x_{4} = \frac{1 + \sqrt{a + 1} + \sqrt{a - 2 + 2 \sqrt{a + 1}}}{2} \in R with a ⩾ 0

btw (1) x_{2} = \frac{1}{x_{1}} \land x_{4} = \frac{1}{x_{3}} because

P_{a} (x) = x^{4} - 2 x^{3} - (a - 2) x^{2} - 2 x + 1 symmetric

(2) no real solution for a < 0 which is easy to see :

P_{a} (x) = 0 \Rightarrow a = \frac{(x^{2} + 1) {(x - 1)}^{2}}{x^{2}} ⩾ 0

Answered by MJS_new last updated on 30/Aug/21

(x^{2} + 1) {(x - 1)}^{2} - {a x}^{2} = 0

x^{4} - 2 x^{3} - (a - 2) x^{2} - 2 x + 1 = 0

x = t + \frac{1}{2}

t^{4} - \frac{2 a - 1}{2} t^{2} - (a + 1) t - \frac{4 a - 5}{16} = 0

(t^{2} - α t - β) (t^{2} + α t - γ) = 0

matching the coefficients leads to

{\begin{cases} α^{2} + β + γ = \frac{2 a - 1}{2} \\ α β - α γ = a + 1 \\ β γ = - \frac{4 a - 5}{16} \end{cases}

solving (1) and (2) for β and γ \Rightarrow

{\begin{cases} β = - \frac{α^{2}}{2} + \frac{a + 1}{2 α} + \frac{2 a - 1}{4} \\ γ = - \frac{α^{2}}{2} - \frac{a + 1}{2 α} + \frac{2 a - 1}{4} \\ α^{6} - (2 a - 1) α^{4} + (a^{2} - 1) α^{2} - {(a + 1)}^{2} = 0 \end{cases}

α = \sqrt{z + \frac{2 a - 1}{3}}

z^{3} - \frac{{(a - 2)}^{2}}{3} + \frac{2 (a + 4) (a^{2} - 10 a - 2)}{27} = 0

testing factors of \frac{2 (a + 4) (a^{2} - 10 a - 2)}{27} \Rightarrow

z = \frac{a + 4}{3} \Rightarrow

α = \sqrt{a + 1} \Rightarrow

β = \frac{\sqrt{a + 1}}{2} - \frac{3}{4} \land γ = - \frac{\sqrt{a + 1}}{2} - \frac{3}{4}

\Rightarrow

(t^{2} - \sqrt{a + 1} t - \frac{2 \sqrt{a + 1} - 3}{4}) (t^{2} + \sqrt{a + 1} t + \frac{2 \sqrt{a + 1} + 3}{4}) = 0

\Rightarrow

(x^{2} - (1 + \sqrt{a + 1}) x + 1) (x^{2} - (1 - \sqrt{a + 1} x + 1) = 0

\Rightarrow

x = \frac{1 + \sqrt{a + 1} \pm \sqrt{a - 2 + 2 \sqrt{a + 1}}}{2} \lor x = \frac{1 + \sqrt{a + 1} \pm \sqrt{a - 2 - 2 \sqrt{a + 1}}}{2}

to get 2 real solutions a ⩾ 0

to get 4 real solutions a ⩾ 8

Answered by EDWIN88 last updated on 01/Sep/21

w e h a v e (x^{2} + 1) (x^{2} - 2 x + 1) - {a x}^{2} = 0

D i v i d i n g b y x^{2} y i e l d s

(x + \frac{1}{x}) (x - 2 + \frac{1}{x}) - a = 0

l e t x + \frac{1}{x} = h \Rightarrow h^{2} - 2 h - a = 0

i t f o l l o w s t h a t h = x + \frac{1}{x} = \frac{2 \pm \sqrt{4 + 4 a}}{2} = 1 \pm \sqrt{a}

w h i c h i n t u r n i m p l i e s t h a t i f

a ⩾ 0 t h e n t h e p o l y n o m i a l

P_{a} (x) h a s r e a l z e r o s

x_{1, 2} = \frac{1 + \sqrt{1 + a} \pm \sqrt{a + 2 \sqrt{1 + a} - 2}}{2}

i n a d d i t i o n i f a ⩾ 8 t h e n P_{a} (x) a l s o

h a s t h e r e a l z e r o s

x_{3, 4} = \frac{1 - \sqrt{1 + a} \pm \sqrt{a - 2 \sqrt{1 + a} - 2}}{2}

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