# Can you please answer this maths question showing complete working out?

 GamerDanFrom: USPosts: 49Votes: 196 Can you please answer this maths question showing complete working out. No online automatic answer type things please. Please show working so I can understand how to answer similar questions. I know you have to first reduce it to a quadratic equation by substituting a letter in for part of the question. The ^ shows that the number is meant to be high up as a power. Solve (x^2-x)^2 + (x^2-x) - 2=0 giving exact values

## Answer No.1

 AnonymousFrom: -Posts: -Votes: - Like you said yourself, we'll set "a" as the variable. a = x^2 - x Now, we have our quadratic equation: a^2 + a - 2 = 0 I used factoring to get this (can also use quadratic formula, mentioned later): (a + 2)(a - 1) = 0 Now we have our solutions for a: a = -2 a = 1 But, that is not our answer yet! We substitute for a: x^2 - x = -2 x^2 - x = 1 Now we have two quadratic equations: x^2 - x + 2 = 0 x^2 - x - 1 = 0 Can't easily factor, so will use quadratic formula to solve: (-b +/- sqrt(b^2 - 4ac)) / 2a First equation gives non-real answers, so from the second equation, exact answers are: (1/2)(1 - sqrt(5)) (1/2)(1 + sqrt(5))

## Answer No.2

 AnonymousFrom: -Posts: -Votes: - Let (x^2-x) = Y then Y^2 + Y^2 = 2 so 2Y^2 = 2 so Y^2 = 1 so Y = sqrt1 = 1 so x^2 - x = 1 so x^2 - x - 1 = 0 There is your quadratic equation (x )(x ) gives you x^2, now you need 2 numbers that add up to 1 and multiply to make 1, you can only get pretty close by using 0.00000000001 and 0.9999999999 so (x + 0.0000000001)(x - 0.99999999999) = 0 so x = -0.0000000001 or x = 0.9999999999 I haven't checked the number of decimal places but they should be the same in each case. This is not a pretty answer so it is probably wrong, but I don't see where

## Answer No.3

 AnonymousFrom: -Posts: -Votes: - Solve (x^2-x)^2 + (x^2-x) - 2=0 giving exact values. Let, . x^2 -x = a a^2 + a - 2 = 0 a^2 +2a - a -2 = 0 a(a +2) -(a +2) = 0 (a - 1)(a + 2) = 0 Thus a = 1, and a = -2 For, a = x^2 -x = 1 ==> x^2 - x - 1 = 0 x = [1 +/- sqrt(5)] / 2 >?==================< ANSWER FOR A = x^2 - x = -2 ==> x^2 - x +2 = 0 x = [1 +/- i.sqrt(7)] / 2 >====================< ANSWER

## Answer No.4

 AnonymousFrom: -Posts: -Votes: - Put x^2-x = t so equation becomes t^2+t-2=0 t^2 + 2t - t - 2 = 0 t(t+2) - 1(t+2) = 0 (t-1)(t+2)=0 therefore t=1 or t = -2 now put value of t i.e. t=x^2-x x^2-x = 1 or x^2-x = -2 x^2-x - 1=0 or x^2-x +2 =0 now solve x^2-x - 1=0 x= (1+sqrt(1+4))/2 by using general formula for solution of ax^2 + bx+c=0 is x=(-b+sqrt(b^2-4ac)/(2a) or x=(-b-sqrt(b^2-4ac)/(2a) or x= (1-sqrt(1+4))/2 now solve x^2-x +2 =0 x= (1+sqrt(1-8))/2 or x= (1-sqrt(1-8))/2 Thus we get 4 roots x= (1+sqrt(5))/2 x= (1-sqrt(5))/2 x= (1+sqrt(7) * i)/2 since sqrt(-1) = i x= (1- sqrt(7) * i)/2

## Answer No.5

 TinyDerived99From: USPosts: 43Votes: 129 (x^2-x)^2 + (x^2-x) -2=0 Rewrite this as ((x-1)^2+1) x^2 = x+2 (-1-x+x^2) (2-x+x^2) =0 Now, a theorem in algebra says ab=0 if a=0 or b=0 Use the quadratic formula x=(-b+sqrt(b^2-4ac))/2a or x=(-b+sqrt(b^2-4ac))/2a. Do this and you will get the solutions.

## Answer No.6

 ZenBrain9From: USPosts: 70Votes: 210 U=x^2-x u^2+u-2=0 (u+2)(u-1)=0 plug u back in (x^2-x+2)(x^2-x-1)=0 (x-2)(x+1)(then quadratic formula this the second part x=2,-1 x^2-x-1=0 1+ (1+4) over 2 = 7/2 1-(1+4) over 2 =-4 x=2,-1,7/2,4

## Answer No.7

 CSHead25From: USPosts: 28Votes: 56 Very simple: let y=x^2 - x so it reduces to y2 + y -2 =0 (y+2) (y-1)=0 y= - 2 or y = 1 so case 1 x2-x = - 2 and x2 -x =1 x2-x+2 =0 and x2 - x -1 =0 x= 1/2 + 1/2 root (-7) x= 1/2 - 1/2 root (-7) and x= 1/2 +1/2 root (-3) x= 1/2 - 1/2 root (-3) Simple

## Answer No.8

 AnonymousFrom: -Posts: -Votes: - Let u = x^2 - x u^2 + u - 2 = 0. (u - 1)(u + 2) = 0 u = 1 and u = -2 x^2 - x - 1 = 0 use the quadratic formula where x = (-b ± √(b^2 - 4ac)) / 2a where ax^2 + bx + c = 0 this gives x = 1.62 and x = -0.62 x^2 - x + 2 = 0 x = 0.5 + 1.32i and x = 0.5 - 1.32ipara: (x^2-x)^2+(x^2-x)-2=0; primero se sustituye (x^2-x) por "a" para que quede como:a^2+a-2=0; luego se despeja: a^2+a=2; se saca factor comun "a": a(a+1)=2; al sustituir "a=1" nos queda: (1)(1+1)=2 2=2. Ahora bien, si a=1 y remplazamos a "a" por x^2-x nos queda: x^2-x=1 y se iguala a 0 con lo que queda: x^2-x-1=0; al utilizar la ecuacion de segundo grado (-b+/-sqrt^2-4*a*c)/2*a nos queda: a=1, b=-1, c=-1 sustituyendo(-(-1)+/-sqrt^2-4*(1)*(-1))/(2*(1)), se soluciona quedando 1+/-sqrt^24/2; se resuelve la raiz quedando 1+/-2/2, siendo la solucion 1. X = 3/2 y 2. X = -1/2

## Answer No.9

 ludwingecharryFrom: VEPosts: 7Votes: 0 Se solucionan las ecuaciones sustituyendo las funciones primero se sustituye (x^2-x) por "a" para que quede como:a^2+a-2=0; luego se despeja: a^2+a=2; se saca factor comun "a": a(a+1)=2; al sustituir "a=1" nos queda: (1)(1+1)=2 2=2. Ahora bien, si a=1 y remplazamos a "a" por x^2-x nos queda: x^2-x=1 y se iguala a 0 con lo que queda: x^2-x-1=0; al utilizar la ecuacion de segundo grado (-b+/-sqrt^2-4*a*c)/2*a nos queda: a=1, b=-1, c=-1 sustituyendo(-(-1)+/-sqrt^2-4*(1)*(-1))/(2*(1)), se soluciona quedandoprimero se sustituye (x^2-x) por "a" para que quede como:a^2+a-2=0; luego se despeja: a^2+a=2; se saca factor comun "a": a(a+1)=2; al sustituir "a=1" nos queda: (1)(1+1)=2 2=2. Ahora bien, si a=1 y remplazamos a "a" por x^2-x nos queda: x^2-x=1 y se iguala a 0 con lo que queda: x^2-x-1=0; al utilizar la ecuacion de segundo grado (-b+/-sqrt^2-4*a*c)/2*a nos queda: a=1, b=-1, c=-1 sustituyendo(-(-1)+/-sqrt^2-4*(1)*(-1))/(2*(1)), se soluciona quedando asi la respuesta cpa

## Answer No.10

 AnonymousFrom: -Posts: -Votes: - Let y=x^2 - x so it reduces to y2 + y -2 =0 (y+2) (y-1)=0 y= - 2 or y = 1 so case 1 x2-x = - 2 and x2 -x =1 x2-x+2 =0 and x2 - x -1 =0 x= 1/2 + 1/2 root (-7) x= 1/2 - 1/2 root (-7) and x= 1/2 +1/2 root (-3) x= 1/2 - 1/2 root (-3) Simple

## Answer No.11

 AnonymousFrom: -Posts: -Votes: - Ut x^2-x = t so equation becomes t^2+t-2=0 t^2 + 2t - t - 2 = 0 t(t+2) - 1(t+2) = 0 (t-1)(t+2)=0 therefore t=1 or t = -2 now put value of t i.e. t=x^2-x x^2-x = 1 or x^2-x = -2 x^2-x - 1=0 or x^2-x +2 =0 now solve x^2-x - 1=0 x= (1+sqrt(1+4))/2 by using general formula for solution of ax^2 + bx+c=0 is x=(-b+sqrt(b^2-4ac)/(2a) or x=(-b-sqrt(b^2-4ac)/(2a) or x= (1-sqrt(1+4))/2 now solve x^2-x +2 =0 x= (1+sqrt(1-8))/2 or x= (1-sqrt(1-8))/2 Thus we get 4 roots x= (1+sqrt(5))/2 x= (1-sqrt(5))/2 x= (1+sqrt(7) * i)/2 since sqrt(-1) = i x= (1- sqrt(7) * i)/2

## Answer No.12

 AnonymousFrom: -Posts: -Votes: - Whatttttttttttt

## Answer No.13

 AnonymousFrom: -Posts: -Votes: - Ma fhemt walo, and thanks

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