# If the sum of the roots of a quadratic equation is 3 and the sum of their cubes is 63 find the equation.

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

Posted on

Let x1 and x2

==> x1+x2 = 3.......(1)

==> x1^3+ x2^3 = 63.......(2)

from (2)

==> x1^3 + x2^3 = (x1+x2)^3 - 3ab(a+b)

==> 63 = 3^3 - 3ab(3)

==> 63 = 27 - 9ab

==> 36 = -9ab

==> ab = 36/-9 = -4

Now we know that the sum = 3

and the product = -6

==> The equation is:

x^2 - 3x - 4 = 0

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll put the roots of the equation as: a and b.

We know that we can from a quadratic equation, knowing the sum and the product of it's roots.

x^2 - Sx + P = 0

S = a+b

P = a*b

We know that the sum of the roots is:

a+b = 3

The sum of their cubes is:

a^3 + b^3 = 63

We could write the sum of the cubes as:

a^3 + b^3 = (a+b)^3 - 3a*b(a+b)

63 = (3)^3 - 3*P*3

63 = 27 - 9*P

We'll subtract 27 both sides:

63-27 = -9P

36 = -9P

We'll divide by -9 both side:

P = -4

Now, knowing the sum and the product of the roots, we can write the equation:

x^2 - 3x - 4 = 0

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Let a and b be the roots of the quadratic equation, then the quadratic equation is  if the form:

x^2 -(a+b)x+ab = 0.....(1).  We have to determine a and b by the given conditions and substitute in in this equation (1).

Given condtions:

a+b =3 .........(2)and

a^3+b^3  = 63.........(3)

We rewrite in  LHS of (3) as:

(a+b)^3-3ab(a+b) = 63. Put a+b = 3 as in (2).

3^3 - 3 ab(3) =63

27 -9ab = 63. Divide by 7.

3-ab =7

ab = 3-7 = -4.

So a+b =3 and ab = -4.

So the required equation is:

x^2-(a+b)x+ab = 0

x^2 -3x - 4 = 0.

thewriter | College Teacher | (Level 1) Valedictorian

Posted on

Assume the roots of the equation we have to find are r1 and r2.

So (x-r1)(x-r2)=0

or x^2-(r1+r2)x+r1*r2=0

Let's write this as x^2-bx+c=0

Here b=r1+r2 and c=r1*r2.

Now it is given that the sum of the roots is 3, therefore b=3.

And as the sum of their cubes is 63, r1^3+r2^3=63.

Now (r1+r2)^3=r1^3+r2^3+3*r1*r2*(r1+r2).

Substituting the values we have, 3^3=63+9*r1*r2.

Therefore r1*r2= (27-63)/9=-4. Or c in the quadratic equation is equal to -4.

The required quadratic equation is x^2-3x-4=0

To check: this quadratic equation has roots 4 and -1. -1+4=3 and (-1)^3+4^3=63.