Solve the following equations: a)(x^2 −a)^2 −6x^2 +4x+2a=0 b)x^4 −4x^3 −10x^3 +37x−14=0,if it known that the left−hand side of the equation can be decomposed into factors with integral coefficients.


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Question Number 113272 by 1549442205PVT last updated on 12/Sep/20

$Solve the following equations :$ $a) {(x^{2} - a)}^{2} - {6 x}^{2} + 4 x + 2 a = 0$ $b) x^{4} - {4 x}^{3} - {10 x}^{3} + 37 x - 14 = 0, if it$ $known that the left - hand side of the$ $equation can be decomposed into$ $factors with integral coefficients .$
	Answered by behi83417@gmail.com last updated on 12/Sep/20
	$(x^2 −a)^2 −4x^2 −2(x^2 −2x−a)=0 (x^2 −2x−a)(x^2 +2x−a)−2(x^2 −2x−a)=0 ⇒(x^2 −2x−a)(x^2 +2x−a−2)=0 ⇒ { ((x^2 −2x−a=0⇒x=((1±(√(1+a)))/2))),((x^2 +2x−a−2=0⇒x=((−1±(√(a+3)))/2))) :}$
	${(x^{2} - a)}^{2} - {4 x}^{2} - 2 (x^{2} - 2 x - a) = 0$ $(x^{2} - 2 x - a) (x^{2} + 2 x - a) - 2 (x^{2} - 2 x - a) = 0$ $\Rightarrow (x^{2} - 2 x - a) (x^{2} + 2 x - a - 2) = 0$ $\Rightarrow {\begin{cases} x^{2} - 2 x - a = 0 \Rightarrow x = \frac{1 \pm \sqrt{1 + a}}{2} \\ x^{2} + 2 x - a - 2 = 0 \Rightarrow x = \frac{- 1 \pm \sqrt{a + 3}}{2} \end{cases}$
	Commented by 1549442205PVT last updated on 12/Sep/20

	$Thank you very much!$
	Answered by 1549442205PVT last updated on 13/Sep/20
	$a)(x^2 −a)^2 −6x^2 +4x+2a=0(∗) ⇔x^4 −2ax^2 +a^2 −6x^2 +4x+2a=0 ⇔a^2 −2(x^2 −1)a+x^4 −6x^2 +4x=0 We look at this like as a quadratic equation with respect to a with the discrimimant Δ′=(x^2 −1)^2 −(x^4 −6x^2 +4x) =4x^2 −4x+1=(2x−1)^2 .Hence, we get a=x^2 −1±(2x−1).From that i)x^2 +2x−2−a=0(1) ii)x^2 −2x−a=0(2) Solving eqns.(1)and (2)with respect to x we ontain:x_(1,2) =1±(√(a+3)) and x_3 ,_4 =1±(√(a+1)) The roots x_(1,2) are real if a∈[−3;+∞) and the roots x_(3,4) are real if a∈[1;+∞) b)We represent the left−hand side of the equation x^4 −4x^3 −10x^3 +37x−14=0 as (x^2 +ax+c)(x^2 +bx+d)=0 ⇔x^4 +(a+b)x^3 +(ab+c+d)x^2 +(bc+ad)x +cd≡x^4 −4x^3 −10x^2 +37x−14.We have a system: { ((a+b=−4)),((ab+c+d=−10)),((bc+ad=37)),((cd=−14)) :} Since a,b,c and d are integers,it follows from the last equation that c=−1,d=14 or c=2,d=−7.The system is completely satisfied by the second pair of values of c and d;for these values we get a=−5 and b=1 for the other coefficients. Solving now the equations x^2 −5x+2=0(1) and x^2 −x−7=0(2) we find the roots of the original equation. x_(1,2) =((5±(√(17)))/2),x_(3,4) =((1±(√(29)))/2).Thus,the given equation has four roots: {((5−(√(17)))/2),((5−(√(17)))/2),((1+(√(29)))/2),((1−(√(29)))/2)}$
	$a) {(x^{2} - a)}^{2} - {6 x}^{2} + 4 x + 2 a = 0 (*)$ $\Leftrightarrow x^{4} - {2 ax}^{2} + a^{2} - {6 x}^{2} + 4 x + 2 a = 0$ $\Leftrightarrow a^{2} - 2 (x^{2} - 1) a + x^{4} - {6 x}^{2} + 4 x = 0$ $We look at this like as a quadratic$ $equation with respect to a with the$ $discrimimant Δ^{'} = {(x^{2} - 1)}^{2} - (x^{4} - {6 x}^{2} + 4 x)$ $= {4 x}^{2} - 4 x + 1 = {(2 x - 1)}^{2} . Hence,$ $we get a = x^{2} - 1 \pm (2 x - 1) . From that$ $i) x^{2} + 2 x - 2 - a = 0 (1)$ $ii) x^{2} - 2 x - a = 0 (2)$ $Solving eqns . (1) and (2) with respect to$ $x we ontain : x_{1, 2} = 1 \pm \sqrt{a + 3} and$ $x_{3},_{4} = 1 \pm \sqrt{a + 1}$ $The roots x_{1, 2} are real if a \in [- 3; + \infty)$ $and the roots x_{3, 4} are real if a \in [1; + \infty)$ $b) We represent the left - hand side of$ $the equation x^{4} - {4 x}^{3} - {10 x}^{3} + 37 x - 14 = 0$ $as (x^{2} + ax + c) (x^{2} + bx + d) = 0$ $\Leftrightarrow x^{4} + (a + b) x^{3} + (ab + c + d) x^{2} + (bc + ad) x$ $+ cd \equiv x^{4} - {4 x}^{3} - {10 x}^{2} + 37 x - 14 . We have$ $a system :$ ${\begin{cases} a + b = - 4 \\ ab + c + d = - 10 \\ bc + ad = 37 \\ cd = - 14 \end{cases}$ $Since a, b, c and d are integers, it follows$ $from the last equation that c = - 1, d = 14$ $or c = 2, d = - 7 . The system is completely$ $satisfied by the second pair of values of$ $c and d; for these values we get a = - 5$ $and b = 1 for the other coefficients .$ $Solving now the equations$ $x^{2} - 5 x + 2 = 0 (1) and x^{2} - x - 7 = 0 (2)$ $we find the roots of the original equation .$ $x_{1, 2} = \frac{5 \pm \sqrt{17}}{2}, x_{3, 4} = \frac{1 \pm \sqrt{29}}{2} . Thus, the$ $given equation has four roots :$ ${\frac{5 - \sqrt{17}}{2}, \frac{5 - \sqrt{17}}{2}, \frac{1 + \sqrt{29}}{2}, \frac{1 - \sqrt{29}}{2}}$
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