Question

Find the roots of the quadratic equation 3y² - 4y+1=0 By
i) completing the square method

ii) the formula​

Answer 1

Answer:

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

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

$y^2 -\frac{4}{3}y +\frac{1}{3}=0$

Now we can complete the square and we got:

$(y^2 -\frac{4}{3}y +\frac{4}{9}) +(\frac{1}{3} -\frac{4}{9})=0$

$(y- \frac{2}{3})^2 =\frac{1}{9}$

We take square root on both sides and we got:

$y-\frac{2}{3}= \pm \frac{1}{3}$

And the solutions for y are:

$y_1 = \frac{1}{3} +\frac{2}{3}=1$

$y_1 = -\frac{1}{3} +\frac{2}{3}=\frac{1}{3}$

ii) $y =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

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

$y =\frac{4 \pm \sqrt{(-4)^2 -4(3)(1)}}{2*3}$

And we got:

$y_1 = 1 , y_2 =\frac{1}{3}$

Step-by-step 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:

$y^2 -\frac{4}{3}y +\frac{1}{3}=0$

Now we can complete the square and we got:

$(y^2 -\frac{4}{3}y +\frac{4}{9}) +(\frac{1}{3} -\frac{4}{9})=0$

$(y- \frac{2}{3})^2 =\frac{1}{9}$

We take square root on both sides and we got:

$y-\frac{2}{3}= \pm \frac{1}{3}$

And the solutions for y are:

$y_1 = \frac{1}{3} +\frac{2}{3}=1$

$y_1 = -\frac{1}{3} +\frac{2}{3}=\frac{1}{3}$

Part ii

We can use the quadratic formula:

$y =\frac{-b \pm \sqrt{b^2 -4ac}}{2a}$

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

$y =\frac{4 \pm \sqrt{(-4)^2 -4(3)(1)}}{2*3}$

And we got:

$y_1 = 1 , y_2 =\frac{1}{3}$