What are the real solutions of $5x^{2} + 3x - 2 = 0$?

Question

Accepted Answer

A. $x = -1$ and $x = \dfrac{2}{5}$

Factor: $5x^{2} + 3x - 2 = (5x - 2)(x + 1)$. Setting the product to zero:
- $5x - 2 = 0 \Rightarrow x = \dfrac{2}{5}$.
- $x + 1 = 0 \Rightarrow x = -1$.
Verify by substitution into $5x^{2} + 3x - 2$:
- At $x = 2/5$: $5 \cdot 4/25 + 3 \cdot 2/5 - 2 = 4/5 + 6/5 - 2 = 10/5 - 2 = 0$ ✓.
- At $x = -1$: $5 - 3 - 2 = 0$ ✓.
Traps B/C/D flip signs or use the wrong factor pair.
Quick check on the constant term: if the roots are $r_{1}, r_{2}$, then $r_{1} \cdot r_{2} = c/a$ for $ax^{2} + bx + c = 0$. Here $c/a = -2/5$, and indeed $(-1) \cdot (2/5) = -2/5$ ✓.

Answer

$x = 1$ and $x = -\dfrac{2}{5}$

Answer

$x = -2$ and $x = 5$

Answer

$x = \dfrac{1}{2}$ and $x = -3$