Question

Use the discriminant to determine the number of real solutions of 3x2+6x−42=0 .(1 point)
real solution(s)

Answer 1

To determine the number of real solutions for the quadratic equation \(3x^2 + 6x - 42 = 0\), we first identify the coefficients \(a\), \(b\), and \(c\):

- \(a = 3\)
- \(b = 6\)
- \(c = -42\)

Next, we use the discriminant formula, which is given by:

\[
D = b^2 - 4ac
\]

Substituting in the values of \(a\), \(b\), and \(c\):

\[
D = 6^2 - 4(3)(-42)
\]

Calculating \(b^2\):

\[
6^2 = 36
\]

Now calculating \(4ac\):

\[
4(3)(-42) = -504
\]

So, the discriminant becomes:

\[
D = 36 - (-504) = 36 + 504 = 540
\]

Now we analyze the value of the discriminant \(D\):

- If \(D > 0\), there are two distinct real solutions.
- If \(D = 0\), there is one real solution (a repeated solution).
- If \(D < 0\), there are no real solutions.

Since \(D = 540\) is greater than 0, we conclude that the equation \(3x^2 + 6x - 42 = 0\) has **two distinct real solutions**.