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 ?

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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.

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william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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




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's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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

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