Question

Find the discriminant of the quadratic equation then state the number and type of solutions 10R^2 +6R=4

Answer 1

To find the discriminant of the quadratic equation, we first need to rewrite the equation in standard form \( ax^2 + bx + c = 0 \).

Starting with the given equation:

\[
10R^2 + 6R - 4 = 0
\]

Here, we can identify:

- \( a = 10 \)
- \( b = 6 \)
- \( c = -4 \)

The discriminant \( D \) of a quadratic equation is given by the formula:

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

Now we can substitute the values of \( a \), \( b \), and \( c \):

\[
D = 6^2 - 4(10)(-4)
\]

Calculating \( D \):

\[
D = 36 - 4(10)(-4)
\]
\[
D = 36 - (-160)
\]
\[
D = 36 + 160
\]
\[
D = 196
\]

The discriminant \( D = 196 \).

Now we can analyze the number and type of solutions based on the value of 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 (the solutions are complex).

Since \( D = 196 \) (which is greater than 0), we conclude that there are **two distinct real solutions** for the equation \( 10R^2 + 6R - 4 = 0 \).