Question

Find the roots of the quadratic equation 3y² - 4y+1=0 By

i) completing the square method

ii) the formula​

Answer 1

i)
`3y^2 -4y +1=0`

We can divide both sides of the equation by 3 and we got:

`y^2 -(4)/(3)y +(1)/(3)=0`

Now we can complete the square and we got:

`(y^2 -(4)/(3)y +(4)/(9)) +((1)/(3) -(4)/(9))=0`

`(y- (2)/(3))^2 =(1)/(9)`

We take square root on both sides and we got:

`y-(2)/(3)= \pm (1)/(3)`

And the solutions for y are:

`y_1 = (1)/(3) +(2)/(3)=1`

`y_1 = -(1)/(3) +(2)/(3)=(1)/(3)`

ii)
`y =(-b \pm √(b^2 -4ac))/(2a)`

And with
`a = 3, b=-4 and c =1` we got:

`y =(4 \pm √((-4)^2 -4(3)(1)))/(2*3)`

And we got:

`y_1 = 1 , y_2 =(1)/(3)`

Explanation:

Part i

For this case we have the following function given:

`3y^2 -4y +1=0`

We can divide both sides of the equation by 3 and we got:

`y^2 -(4)/(3)y +(1)/(3)=0`

Now we can complete the square and we got:

`(y^2 -(4)/(3)y +(4)/(9)) +((1)/(3) -(4)/(9))=0`

`(y- (2)/(3))^2 =(1)/(9)`

We take square root on both sides and we got:

`y-(2)/(3)= \pm (1)/(3)`

And the solutions for y are:

`y_1 = (1)/(3) +(2)/(3)=1`

`y_1 = -(1)/(3) +(2)/(3)=(1)/(3)`

Part ii

We can use the quadratic formula:

`y =(-b \pm √(b^2 -4ac))/(2a)`

And with
`a = 3, b=-4 and c =1` we got:

`y =(4 \pm √((-4)^2 -4(3)(1)))/(2*3)`

And we got:

`y_1 = 1 , y_2 =(1)/(3)`