Question

3y2-5y+2=0, using the completing the square method

Answer 1

Answer:

3y^2-5y+2=0

a = 3; b = -5; c = +2;

Δ = b2-4ac

Δ = -52-4·3·2

Δ = 1

The delta value is higher than zero, so the equation has two solutions

We use following formulas to calculate our solutions:

y1=−b−Δ√2ay2=−b+Δ√2a

Δ−−√=1√=1

y1=−b−Δ√2a=−(−5)−12∗3=46=2/3

y2=−b+Δ√2a=−(−5)+12∗3=66=1

Answer 2

Answer:

(1): "y2" was replaced by "y^2".

Step by step solution :

STEP

1

:

Equation at the end of step 1

(3y2 + 5y) - 2 = 0

STEP

2

:

Trying to factor by splitting the middle term

2.1 Factoring 3y2+5y-2

The first term is, 3y2 its coefficient is 3 .

The middle term is, +5y its coefficient is 5 .

The last term, "the constant", is -2

Step-1 : Multiply the coefficient of the first term by the constant 3 • -2 = -6

Step-2 : Find two factors of -6 whose sum equals the coefficient of the middle term, which is 5 .

-6 + 1 = -5

-3 + 2 = -1

-2 + 3 = 1

-1 + 6 = 5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -1 and 6

3y2 - 1y + 6y - 2