How do you solve the quadratic $3 y^{2} - y - 4 = 0$ $3y^2-y-4=0$ using any method?

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Answer 1 · 2016-10-07T10:29:08

$y = - 1 or y = \frac{4}{3}$ $y=-1 or y = 4/3$

Find factors of 3 and 4 which subtract to give 1.

Notice that the numbers themselves are the required factors because $4 - 3 = 1$ $4-3=1$

$(3 y - 4) (y + 1) = 0$ $(3y-4)(y+1) = 0$

Either of the 2 brackets could be equal to 1.

If $3 y - 4 = 0 \to y = \frac{4}{3}$ $3y -4 = 0 rarr y =4/3$

If $y + 1 = 0 \to y = - 1$ $y+1 =0 rarr y = -1$

How do you solve the quadratic 3y2−y−4=03y^2-y-4=0 using any method?