how do i solve this quadratic function x^2-6x+8=0 what i did was i factored it like : (x-4) (x-2) x=4 x=2 is this correct ?
you did correctly.
The given quadratic equation is in the form x^2-6x+8 = 0.
So we factor the left side and apply the zero prodict rule by setting each factor to zero.
x^2-6x+8 = 0.
The LHS cuold be grouped as below to factor:
x^2-4x -2x+8 = 0
x(x-4) -2(x-4) = 0.
(x-4)(x-2) = 0.
Apply the zero product rule that if ab = 0, then a= 0 or b = 0.
x-4 = 0 or x-2 = 0.
x-4 = 0 gives x= 4.
x-2 = 0 gives x = 2.
We have to solve x^2-6x+8=0
Now we express -6x as two terms such that their product is 8x^2,
that would be -4x and -2x
As you got the same factors you were correct.
Further (x-2)(x-4)=0 gives us that x-2 or x-4 is equal to zero.
Therefore x=2 and x=4
To verify if the roots are correctly calculated, we'll use Viete's relations.
Because we'll need the coefficients of the quadratic, first, we'll write the equation in the general form:
ax^2 + bx + c = 0
We'll identify the coefficients a,b,c:
a = 1
b = -6
c = 8
These relations link the roots of the equation and the coefficients of the equation, in this way:
x1 + x2 = -b/a
The sum of the roots of the equation is the ratio of the coefficients b and a.
x1*x2 = c/a
The product of the roots is the ratio of the coefficients c and a.
We'll substitute the coefficients a,b,c and the calculated roots and we'll verify the identities:
x1+x2 = -(-6)/1
x1+x2 = 6
We'll substitute x1 by 2 and x2 by 4:
2+4 = 6, true!
x1*x2 = 8
2*4 = 8
Since both Viete's relations have been verified, the calculated roots are valid.
x1 = 2
x2 = 4