# Determine a if the roots of the equation x^2-x-a=0 are x1 and x2 and x1^4+x2^4=1.

justaguide | College Teacher | (Level 2) Distinguished Educator

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The roots of the equation x^2 - x - a = 0 are

x1 = 1 / 2 + sqrt ( 1 + 4a)/2

x2 = 1/2 - sqrt ( 1+ 4a) / 2

Also x1^4 + x2^4 = 1

=> (1 / 2 + sqrt ( 1 + 4a)/2)^4 + (1 / 2 + sqrt ( 1 + 4a)/2)^4 = 1

=> 1/16[(1 + sqrt (1+4a))^4 + (1 - sqrt (1+4a))^4] = 1

now we use a^2 - b^2 = (a-b)(a+b)

=> ((1 + sqrt (1+4a))^2 + (1 - sqrt (1+4a))^2)*((1 + sqrt (1+4a))^2 -(1 - sqrt (1+4a))^2) = 16

=> ((1 + sqrt (1+4a))^2 + (1 - sqrt (1+4a))^2)*(1 + sqrt (1+4a)) + (1 - sqrt (1+4a))(1 + sqrt (1+4a)) - (1 - sqrt (1+4a)) = 16

=> [1 + 1 + 4a + 2*sqrt (1+4a)+1 + 1+4a - 2*sqrt(1+4a)]*[2*2*sqrt (1+4a)] = 16

=> [4 + 8a ]*[4*sqrt (1+4a) = 16

=> (1+ 2a) * sqrt (1+ 4a) = 1

=> (1+ 4a^2 + 4a)(1+ 4a) = 1

=> 1+ 4a + 4a^2 + 16a^3 + 4a + 16a^2 = 1

=> 8a + 20a^2 + 16a^3 = 0

=> 16a^3 + 20a^2 + 8a = 0

=> 4a^3 + 5a^2 + 2a = 0

=> a( 4a^2 + 5a + 2)=0

a1 = 0

a2 = -5/8 + sqrt(25 - 32)/8

=> a2 = -5/8 +i* sqrt 7 / 8

a3 = -5/8 - sqrt 7/8

Therefore a can be 0 , -5/8 +i* sqrt 7 / 8 and -5/8 - sqrt 7/8

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We can find x1^4 + x2^4 developing the binomial:

(x1^2 + x2^2)^2 = x1^4 + x2^4 + 2(x1*x2)^2

We'll subtract 2(x1*x2)^2 both sides:

x1^4 + x2^4 = (x1^2 + x2^2)^2 - 2(x1*x2)^2 (1)

We can determine the sum of the squares of the roots from Viete's relations:

(x1 + x2)^2 = x1^2 + x2^2 + 2x1*x2

x1^2 + x2^2 = (x1 + x2)^2  - 2x1*x2

x1 + x2 = 1

x1*x2 = -a (2)

x1^2 + x2^2 = 1 - 2*(-a)

x1^2 + x2^2 = 1 + 2a (3)

We'll substitute (2) and (3) in (1):

x1^4 + x2^4 = (1 + 2a)^2 - 2(-a)^2

We'll expand the square:

x1^4 + x2^4 = 1 + 4a +  4a^2 - 2a^2

We'll combine like terms:

x1^4 + x2^4 = 2a^2 + 4a + 1

We know, from enunciation, that:

x1^4 + x2^4 = 1

2a^2 + 4a + 1 = 1

We'll subtract 1:

2a^2 + 4a = 0

We'll factorize by 2a:

2a(a + 2) = 0

We'll set the first factor as zero:

2a = 0

a = 0

We'll set the next factor as zero:

a + 2 = 0

a = -2

The possible real values for a are: {-2 ; 0}.

neela | High School Teacher | (Level 3) Valedictorian

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To determine x^2-x-a = 0,

x1 and x2 are the roots of the equation.

Therefore by the relation between the roots and coefficients we have x1+x2 = -1.

x1x2 = a.

Therefore x^4+x4 = (x1^2+x2^2)^2 - 2(x1x2)^2

= {(x1+x2)^2-2x1x2}^2 - 2(x1x2)^2

= {(-1)^2 -2a}^2 - 2(a)^2.

= {1-2a}^2 - 2a^2

= 1-4a+4a^2 -2a^2 .

= 1-4a+2a^2 .

Therefore x1^4+x2^4 = 1 = 1-4a+2a^2.

=> 1= 1-4a+2a^2

=> 0 =-4a+2a^2.

Or 2a(-2+a) = 0.

a= 0, or -2+a = 0.

a= 0, or a = 2.

Therefore the value of a = 0, or a = 2.