# What are the roots of the quadratic equation if the sum is 5 and the product is 6 ?

*print*Print*list*Cite

### 4 Answers

Let f(x) = ax^2 + bx + c is a quadratic function such that x1 and x2 are the roots.

==> Given that:

x1 + x2 = 5

x1*x2 = 6

But we know that:

x1+ x2 = -b/a = 5 ==> b= -5 a

Also, we know that:

x1*x2 = c/a = 6 ==> c = 6 a

==> f(x) = ax^2 -5a x + 6a

We need to find the roots.

==> ax^2 - 5a x + 6 a = 0

We will divide by a:

==> x^2 -5x + 6 = 0

==> (x -2)(x-3) = 0

**Then, the roots are : x = { 2, 3}**

We have the sum of the roots of the quadratic equation as 5 and the product of the roots as 6.

Let the roots be A and B. Here we don't need to consider the quadratic equation. We can find A and B just by using the fact that A*x2 = 6 and A + B = 5

A + B = 5

=> A = 5 - B

substituting this in A*B = 6

=> B*( 5 - B) = 6

=> 5*B - B^2 = 6

=> B^2 - 5B + 6 = 0

=> B^2 - 3B - 2B + 6 =0

=> B(B - 3) - 2(B -3) = 0

=> (B - 2)(B -3) =0

B = 2 and 3

A = 3 and 2

**The required roots are 2 and 3.**

The sum of the roots = 5 and product of the roots is 6.

To find the roots.

So the required quadratic equation x^2 - (sum of the roots)x+ product of the roots = 0.

=> x^2-5x+6 = 0.

=> (x-3)(x-2) = 0.

=> x1 = 3 and x2 = 2.

So the roots of the quadratic equation are x1 = 3 and x2 = 2.

Tally: Sum of the roots = x1+x2 = 3+2 = 5.

Product of the roots = x1x2 = 3*2 = 6.

We know from enunciation that S = 5 and P = 6.

We'll form the quadratic equation knowing the sum and the product:

x^2 - Sx + P = 0

x^2 - 5x + 6 = 0

We'll apply the quadratic formula:

x1 = [5+sqrt(25 - 24)]/2

x1 = (5 + 1)/2

x1 = 3

x2 = (5 - 1)/2

x2 = 2

**The roots of the quadratic equation are {2 ; 3}.**