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 ?

neela | Student

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.

william1941 | Student

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

Therefore x^2-6x+8=0

=>x^2-4x-2x+8=0

=>x(x-4)-2(x-4)=0

=>(x-2)(x-4)=0.

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

giorgiana1976 | Student

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