Which is the value of the sum x1^3 + x2^3 if x1,x2 are solutions of the equation x^2-5x+6=0

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hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

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f(x) = x^2 - 5x + 6

x1 and x2 are solution

We need to determine the value of x1^3 + x2^3

From viete's rule, we know that:

x1+ x2 = -b/a = 5.........(1)

x1*x2 = c/a = 6...........(2)

But f(x1) = f(x2) =0

==> f(x1) = x1^2 -5x1 + 6 = 0 ......(3)

==> f(x2) = x2^2 - 5x2 + 6 =0 ...(4)

Multiply 3 by x1 and (4) by x2

==> x1^3 - 5x1^2 + 6x1 = 0

==> x2^3 - 5x2^2 + 6x2 = 0

Add both equations:

==> x1^3 + x2^3 = 5x1^2 + 5x2^2 - 6x1 -6x2

==> x1^3 + x2^3 = 5(x1^2 + x2^2) - 6(x1+x2)

                             = 5[(x1+x2)^2 - 2x1*x2] - 6(x1+x2)

                            = 5(5^2) - 5(2*6) - 6(5)

                            = 125 - 60 - 30

                            =125 - 90 = 35

==> x1^3 + x2^3 = 35

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neela | High School Teacher | (Level 3) Valedictorian

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If x1 and x2 are the solutions of x^2-5x+6 =0 .

The solutions are of equation(x-2)(x-3) = 0 or solution x1 =2 and x2 = 3 and  x1^3+x2^3 = 2^3+3^3 = 8+27 = 35.

Also  by the reelation between roots and coefficients,

x1+x2 = - -5/1 = 5...........(1)

x1*x2 = 6/1 = 6...............(2)

We know thw identity

 x1^3+x2^3= (x1+x2)^3 - 3x1x2(x1+x2) ..........(3)

 Substitite the values of x1+x2 and x1x2 from (1) and (2) in  (3) :

X1#+x2^3 = (5)^3-3*6(-5) = 125-90 = 35.

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thewriter | College Teacher | (Level 1) Valedictorian

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We have the quadratic equation x^2-5x+6=0

To solve this: x^2-5x+6=x^2-3x-2x+6=x(x-3)-2(x-3)=(x-2)(x-3)=0

The roots of the equation are derived by equating x-2 and x-3 to 0, doing which we get x=2 and x=3 resp.

As x1 and x2 are the solutions of the the equation we just solved x1=2 and x2=3. (It does not make a difference whether x1 and x2 are assigned 2 or 3 as both have been raised to the power 3 in the expression we have to find the value of)

The value of x1^3+x2^3=2^3+3^3=8+27=35

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giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate the sum of the cubes of the solutions of the equation, we'll use Viete's relationships.

x1+x2=-b/a

x1*x2=c/a, where a,b,c are the coefficients of the equation:

ax^2+bx+c=0, in our case:a=1,b=-5 and c=6.

x1+x2=-(-5/1)=5

x1*x2=6

For calculating x1^3+x2^3, we'll put each solution into equation:

x1^2-5x1+6=0  (1)

x2^2-5x2+6=0  (2)

We'll multiply (1) with x1 and (2) with x2.

x1^3-5x1^2+6x1=0 (3)

x2^3-5x2^2+6x2=0  (4)

We'll add (3)+(4):

(x1^3+x2^3)-5(x1^2+x2^2)+6(x1+x2)=0

x1^2+x2^2=(x1+x2)^2 - 2x1*x2=5^2-12=25-12=13

(x1^3+x2^3) = 5(x1^2+x2^2)-6(x1+x2)

(x1^3+x2^3) = 5*13 - 6*5 = 65-30

S=(x1^3+x2^3)=35

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