Question

Solve equation by factoring
2x^2+5x-3=0

Answer 1

To solve the quadratic equation \( 2x^2 + 5x - 3 = 0 \) by factoring, we can start by looking for two numbers that multiply to \( a \cdot c \) (where \( a \) is the coefficient of \( x^2 \) and \( c \) is the constant term) and add up to \( b \) (the coefficient of \( x \)).

In this case:
- \( a = 2 \)
- \( b = 5 \)
- \( c = -3 \)

First, we calculate \( a \cdot c = 2 \cdot (-3) = -6 \).

Now we need two numbers that multiply to \(-6\) and add to \(5\). The correct numbers are \(6\) and \(-1\) because:
- \(6 \cdot (-1) = -6\)
- \(6 + (-1) = 5\)

Now we can rewrite the middle term \(5x\) using these two numbers:
\[
2x^2 + 6x - 1x - 3 = 0
\]

Next, we can factor by grouping:
\[
(2x^2 + 6x) + (-1x - 3) = 0
\]

Factoring out the greatest common factors from each group gives us:
\[
2x(x + 3) - 1(x + 3) = 0
\]

Now we can factor out the common term \((x + 3)\):
\[
(2x - 1)(x + 3) = 0
\]

Setting each factor equal to zero gives us two equations to solve:
1. \( 2x - 1 = 0 \)
2. \( x + 3 = 0 \)

Solving these equations:

For \( 2x - 1 = 0 \):
\[
2x = 1 \\
x = \frac{1}{2}
\]

For \( x + 3 = 0 \):
\[
x = -3
\]

Thus, the solutions to the equation \( 2x^2 + 5x - 3 = 0 \) are:
\[
x = \frac{1}{2} \quad \text{and} \quad x = -3
\]