Question

Use the discriminant to determine the number of real solutions of 3x2+6x−42=0
3
𝑥
2
+
6
𝑥
−
42
=
0
.

Answer 1

To determine the number of real solutions of the quadratic equation \(3x^2 + 6x - 42 = 0\), we can use the discriminant. The discriminant \(D\) is given by the formula:

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

where \(a\), \(b\), and \(c\) are the coefficients of the quadratic equation in the standard form \(ax^2 + bx + c = 0\).

For the given equation:

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

Now, we can calculate the discriminant:

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

Calculating \(6^2\):

\[
6^2 = 36
\]

Now calculating \(4(3)(-42)\):

\[
4 \times 3 \times -42 = -504
\]

So, we have:

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

Now, we analyze the discriminant:

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

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