Question

Select the two values of x that are roots of this equation.
2x- 5=-3x2
0 x= 3
0 x=-5
0 x= 1
0 x=-1/2

Answer 1

To solve the equation \( 2x - 5 = -3x^2 \), we first rearrange it into standard polynomial form:

\[
3x^2 + 2x - 5 = 0
\]

Next, we can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = 2 \), and \( c = -5 \).

First, we calculate the discriminant \( b^2 - 4ac \):

\[
b^2 - 4ac = 2^2 - 4(3)(-5) = 4 + 60 = 64
\]

Now we apply the quadratic formula:

\[
x = \frac{-2 \pm \sqrt{64}}{2 \cdot 3}
\]

\[
x = \frac{-2 \pm 8}{6}
\]

This gives us two possible values for \( x \):

1. \( x = \frac{-2 + 8}{6} = \frac{6}{6} = 1 \)

2. \( x = \frac{-2 - 8}{6} = \frac{-10}{6} = -\frac{5}{3} \)

From the options given (3, -5, 1, -1/2), the value \( x = 1 \) is clearly among the options. However, \( -\frac{5}{3} \) is not included in your choices.

Since \( 3 \) and \( -5 \) are not solutions (as they do not satisfy the quadratic equation), and neither does \( -1/2 \), the only root from your options is:

\[
\boxed{1}
\]

You provided an incomplete set of values to choose from since we found \( -\frac{5}{3} \) as the second root. Thus, the roots of the equation based on what you provided are:

1. \( x = 1 \)

For further clarity, if any of these provided options correspond to roots, it needs to be checked against the obtained \( -\frac{5}{3} \). In essence, from the list provided, the only valid root is \( x = 1 \).