Simplify equation with square and non-square $y$

Question

I'm trying to solve a few equations but my math is a bit (very) rusty. In particular, I've simplified one of them to this:

$$2y^2-3y-2=0$$

I can see that the answer is "2" but how can I prove it? Is it possible to simplify the equation further?

Answer 1

Here's a nice way to factor a quadratic $ax^2 + bx+ c$. First find two numbers whose product is $ac$ and whose sum is $b$. Next rewrite $b$ as this sum, and then group the terms in pairs and factor accordingly.

In your case, with $2y^2 - 3y - 2$, we need two numbers whose product is $-4$ and whose sum is $-3$; the numbers must be $-4$ and $1$. So we rewrite the term $-3y$ as $-4y + y$, giving the following: $$2y^2 - 4y + y -2 = 0$$ Now look at the terms in pairs and pull out common factors: $$(2y^2 - 4y) + (y -2) = 0\\ 2y(y - 2) + (y -2) = 0$$ You can see that there's now a common term of $y-2$, so we can factor again: $$(2y+1)(y-2) = 0,$$ whence $y = -\frac12$ or $y = 2$.

Answer 2

Hint. You may use the formula for the roots of second degree polynomials. If you have a polynomial of degree two, say $ax^2 +bx+c$, the roots are given by $$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$

Answer 3

You've stated an answer. There may be another distinct answer.

Pulling out the trusty quadratic formula:

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

$$y = \frac{3 \pm 5}{4}$$

$$y = 2 \text{ or } -1/2$$