# Which is the sum of the cubes of the roots of equation x^2+ax-4=0?

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

Posted on

f(x)= x^2+ax-4

Assume that the roots are x1 and x2

then:

x1^2+ax1-4=0 ....(1)

x2^2+ax2-4=0 ....(2)

==> x1^2+x2^2+ax1+ax2-8=0

==> x1^2+x2^2= 8-a(x1+x2)

but (x1+x2)= -a

==> x1^2+x2^2=8+a^2......(3)

Now we need to calculate the sum of the cubes of the roots:

that means we need x1^3+x2^3

Let us multiply equation (1) with x1 and equation (2) with x2

==> x1^3+ax1^2-4x1=0 .......(4)

==> x2^3+ax2^2-4x2=0 ........(5)

Now let us add (4) and (5)

==> x1^3+ax1^2-4x1+x2^3+ax2^2-4x2=0

==> x1^3+x2^3 = 4x1+4x2-ax1^2-ax2^2

factor the right side:

==> x1^3+x2^3 = 4(x1+x2)-a(x1^2+x2^2)

But from equation (3) we have x1^2+x2^2=8+a^2

==> x1^3+X2^3 = 4(-a)-a(8+a^2)

= -4a-8a-a^3

= -12-a^3

Then  x1^3+x2^3= -12-a^3

neela | High School Teacher | (Level 3) Valedictorian

Posted on

x^2+ax-4 = 0. To find the sum of the cubes of the solution.

If x1 ans x2 are the solutions of this quadratic equation then

x1+x2 = -a and x1*x2 = -4 by the relation between coeficient and roots

x1^2+ax1-4 = 0.....(a)  and x2^2+ax2-4 = 0.... (b). Adding the two equations, (x1^2+x2^2) +a(x1+x2) - 8 = 0. Or x1^2+x2^2 = -a(x1+x2)+8 = -a(-a) +8 = a^2+8

Multiplying eq(a) by x1 , we get:

X1^3+ax1^2-4x1 = 0. ...........(1)Similarly,

x2^3+ax2^2-4x2 = 0..............(2)

(1)+(2) gives:

(x1^3+x2^3 )+a(x1^2+x2^2) -4(x1+x2) = 0. Or

x1^3+x2^3 = -a(x1^2+x2^2)+4(x1+x2)

= -a(a^2+8) +4(-a) = -a^3-12a

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Let's remember the fact that a solution of an equation, substituted into equation, verifies it.

Let's substitute the first root into equation and we'll get:

x1^2+ax1-4=0 (1)

Let's substitute the second root into equation:

x2^2+ax2-4=0  (2)

Now, we'll multiply (1) by x1 and (2) by x2 and we'll get:

x1^3+ax1^2-4x1=0 (3)

x2^3+ax2^2-4x2=0  (4)

Now. we'll add the relations (3) and (4):

x1^3+x2^3 = -a(x1^2+x2^2)+4(x1+x2) (5)

We'll find the value of (x1 + x2) and x1*x2, using the Viete's relation:

x1+x2 = - a

x1*x2 = 2

We'll substitute (x1+x2) by -a and

x1^2 + x2^2 = (x1+x2)^2-2x1*x2 = a^2-4, into the relation (5) and we'll obtain:

x1^3+x2^3 = -a(a^2-4)+4(-a)

x1^3+x2^3 = -a^3+4a-4a

Aftyer reducing similar terms, we'll obtain:

x1^3+x2^3 = -a^3