Question

B) Find the root of quad equation if they exist
by method of completing the square
gy
$9y { }^{2} - 12y + 2 = 0$
​

Answer 1

Given :-9y^2-12y+2=0

Now divide 9 on both the sides so as to bring a=1 =>9y^2/9-12y/9+2/9=0/9 =>y^2-4/3y+2/9=0 Now take 2/9 to the right hand side =>y^2-4/3y=-2/9 Now add (2/3)^2 on both the sides =>(y)^2-2*x*2/3+(2/3)^2=(-2/9)+(2/3)^2 =>(y-2/3)^2=-2/9+4/9 =>(y-2/3)^2=-2+4/9 =>(y-2/3)^2=2/9 =>y-2/3=+- √2/9 =>y-2/3=+- √2/3 =>y=√2/3+2/3=√2+2/3 =>y=-√2/3+2/3=-√2+2/3 [This is the required answer] [If you like my answer then please mark me as brainliest and I will follow you back.]

Answer 2

Answer:

fraction answers: y = $\frac{\sqrt{2}+2}{3}$ and y = $-\frac{\sqrt{2}+2}{3}$

rounded answers:   y ≈ 1.138 and y ≈ -1.138

Step-by-step explanation:

starting equation:

$9y^{2}-12y+2=0$

completed square:

$(3y-2)^{2}-2=0$

algebra:

$(3y-2)^{2}=2\\ 3y-2=\sqrt{2}\\ 3y=\sqrt{2}+2\\ y=\frac{\sqrt{2}+2}{3}$

answers as a fraction:

$y=\frac{\sqrt{2}+2}{3}$

and

$y=-\frac{\sqrt{2}+2}{3}$

answers when rounded to nearest thousandth:

y = 1.138

y = -1.138