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


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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
	$(x^2 +1)(x−1)^2 −ax^2 =0 x^4 −2x^3 −(a−2)x^2 −2x+1=0 x=t+(1/2) t^4 −((2a−1)/2)t^2 −(a+1)t−((4a−5)/(16))=0 (t^2 −αt−β)(t^2 +αt−γ)=0 matching the coefficients leads to { ((α^2 +β+γ=((2a−1)/2))),((αβ−αγ=a+1)),((βγ=−((4a−5)/(16)))) :} solving (1) and (2) for β and γ ⇒ { ((β=−(α^2 /2)+((a+1)/(2α))+((2a−1)/4))),((γ=−(α^2 /2)−((a+1)/(2α))+((2a−1)/4))),((α^6 −(2a−1)α^4 +(a^2 −1)α^2 −(a+1)^2 =0)) :} α=(√(z+((2a−1)/3))) z^3 −(((a−2)^2 )/3)+((2(a+4)(a^2 −10a−2))/(27))=0 testing factors of ((2(a+4)(a^2 −10a−2))/(27)) ⇒ z=((a+4)/3) ⇒ α=(√(a+1)) ⇒ β=((√(a+1))/2)−(3/4)∧γ=−((√(a+1))/2)−(3/4) ⇒ (t^2 −(√(a+1))t−((2(√(a+1))−3)/4))(t^2 +(√(a+1))t+((2(√(a+1))+3)/4))=0 ⇒ (x^2 −(1+(√(a+1)))x+1)(x^2 −(1−(√(a+1))x+1)=0 ⇒ x=((1+(√(a+1))±(√(a−2+2(√(a+1)))))/2)∨x=((1+(√(a+1))±(√(a−2−2(√(a+1)))))/2) to get 2 real solutions a≥0 to get 4 real solutions a≥8$
	$(x^{2} + 1) {(x - 1)}^{2} - {a x}^{2} = 0$ $x^{4} - 2 x^{3} - (a - 2) x^{2} - 2 x + 1 = 0$ $x = t + \frac{1}{2}$ $t^{4} - \frac{2 a - 1}{2} t^{2} - (a + 1) t - \frac{4 a - 5}{16} = 0$ $(t^{2} - α t - β) (t^{2} + α t - γ) = 0$ $matching the coefficients leads to$ ${\begin{cases} α^{2} + β + γ = \frac{2 a - 1}{2} \\ α β - α γ = a + 1 \\ β γ = - \frac{4 a - 5}{16} \end{cases}$ $solving (1) and (2) for β and γ \Rightarrow$ ${\begin{cases} β = - \frac{α^{2}}{2} + \frac{a + 1}{2 α} + \frac{2 a - 1}{4} \\ γ = - \frac{α^{2}}{2} - \frac{a + 1}{2 α} + \frac{2 a - 1}{4} \\ α^{6} - (2 a - 1) α^{4} + (a^{2} - 1) α^{2} - {(a + 1)}^{2} = 0 \end{cases}$ $α = \sqrt{z + \frac{2 a - 1}{3}}$ $z^{3} - \frac{{(a - 2)}^{2}}{3} + \frac{2 (a + 4) (a^{2} - 10 a - 2)}{27} = 0$ $testing factors of \frac{2 (a + 4) (a^{2} - 10 a - 2)}{27} \Rightarrow$ $z = \frac{a + 4}{3} \Rightarrow$ $α = \sqrt{a + 1} \Rightarrow$ $β = \frac{\sqrt{a + 1}}{2} - \frac{3}{4} \land γ = - \frac{\sqrt{a + 1}}{2} - \frac{3}{4}$ $\Rightarrow$ $(t^{2} - \sqrt{a + 1} t - \frac{2 \sqrt{a + 1} - 3}{4}) (t^{2} + \sqrt{a + 1} t + \frac{2 \sqrt{a + 1} + 3}{4}) = 0$ $\Rightarrow$ $(x^{2} - (1 + \sqrt{a + 1}) x + 1) (x^{2} - (1 - \sqrt{a + 1} x + 1) = 0$ $\Rightarrow$ $x = \frac{1 + \sqrt{a + 1} \pm \sqrt{a - 2 + 2 \sqrt{a + 1}}}{2} \lor x = \frac{1 + \sqrt{a + 1} \pm \sqrt{a - 2 - 2 \sqrt{a + 1}}}{2}$ $to get 2 real solutions a ⩾ 0$ $to get 4 real solutions a ⩾ 8$
	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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