Solving a quadratic equation by factoring and checking roots

Expert intro: This problem tests whether you can factor a quadratic polynomial, set each factor equal to zero, and verify the solution pair against the answer choices.

Detailed walkthrough

Identify the quadratic structure

We want to solve

$2x^{2} + 3x - 2 = 0$.

A standard first step is to factor the quadratic into two binomials. Since the leading coefficient is 2 and the constant term is -2, we look for a factorization of the form

$(2x \pm 1)(x \pm 2)$.

Factor the expression

Try expanding $(2x - 1)(x + 2)$:

$2x^2 + 4x - x - 2 = 2x^2 + 3x - 2$.

That matches exactly, so

$2x^2 + 3x - 2 = (2x - 1)(x + 2)$.

Solve each factor

Set each factor equal to zero:

$2x - 1 = 0 \Rightarrow x = \frac{1}{2}$

$x + 2 = 0 \Rightarrow x = -2$

So the solution set is

$\boxed{x = \frac{1}{2} \text{ and } x = -2}$.

Match the answer choice

That corresponds to choice (b).

Common algebra check

A quick verification is useful: multiply the roots back into the factor form, and confirm the signs. The middle term must be positive, so the factors should produce one positive and one negative middle contribution. That is exactly what happens here.

💡 Pitfall guide

A common mistake is choosing roots that make the signs look convenient without actually factoring the trinomial. Another frequent error is forgetting that the constant term is negative, so the factors must have opposite signs. If you expand a guessed factorization and it does not return $2x^2+3x-2$, the answer is not correct. Always check by substitution or by re-expansion before selecting the choice.

🔄 Real-world variant

If the equation were changed to $2x^2 + 3x + 2 = 0$, the factoring pattern would no longer give real roots in the same way because the constant term is positive. In that variant, you would test factor pairs of $2\cdot 2 = 4$ and look for two numbers that add to 3, but none work with the same sign pattern. A similar change like $2x^2 - 3x - 2 = 0$ would instead factor as $(2x+1)(x-2)=0$, giving roots $x=-\frac12$ and $x=2$.

🔍 Related terms

factoring quadratics, zero product property, trinomial factorization