Question

1. $5n^2 = -4 + 9n$ 10) $3r^2 - 2 = r$

Explanation:

Problem 9: \( 5n^2 = -4 + 9n \)

Step 1: Rearrange into standard quadratic form

A quadratic equation is in the form \( ax^2 + bx + c = 0 \). So, we rearrange \( 5n^2 = -4 + 9n \) to \( 5n^2 - 9n + 4 = 0 \). Here, \( a = 5 \), \( b = -9 \), \( c = 4 \).

Step 2: Use the quadratic formula or factor

Let's try factoring. We need two numbers that multiply to \( a \times c = 5 \times 4 = 20 \) and add up to \( b = -9 \). The numbers are -4 and -5. So we rewrite the middle term:
\( 5n^2 - 4n - 5n + 4 = 0 \)
Group the terms:
\( (5n^2 - 4n) + (-5n + 4) = 0 \)
Factor out the common factors from each group:
\( n(5n - 4) - 1(5n - 4) = 0 \)
Now, factor out \( (5n - 4) \):
\( (5n - 4)(n - 1) = 0 \)

Step 3: Solve for \( n \)

Set each factor equal to zero:
\( 5n - 4 = 0 \) or \( n - 1 = 0 \)
For \( 5n - 4 = 0 \), add 4 to both sides: \( 5n = 4 \), then divide by 5: \( n = \frac{4}{5} \)
For \( n - 1 = 0 \), add 1 to both sides: \( n = 1 \)

Problem 10: \( 3r^2 - 2 = r \)

Step 1: Rearrange into standard quadratic form

Rearrange \( 3r^2 - 2 = r \) to \( 3r^2 - r - 2 = 0 \). Here, \( a = 3 \), \( b = -1 \), \( c = -2 \).

Step 2: Factor the quadratic

We need two numbers that multiply to \( a \times c = 3 \times (-2) = -6 \) and add up to \( b = -1 \). The numbers are -2 and 3. Rewrite the middle term:
\( 3r^2 + 2r - 3r - 2 = 0 \)
Group the terms:
\( (3r^2 + 2r) + (-3r - 2) = 0 \)
Factor out the common factors:
\( r(3r + 2) - 1(3r + 2) = 0 \)
Factor out \( (3r + 2) \):
\( (3r + 2)(r - 1) = 0 \)

Step 3: Solve for \( r \)

Set each factor equal to zero:
\( 3r + 2 = 0 \) or \( r - 1 = 0 \)
For \( 3r + 2 = 0 \), subtract 2 from both sides: \( 3r = -2 \), then divide by 3: \( r = -\frac{2}{3} \)
For \( r - 1 = 0 \), add 1 to both sides: \( r = 1 \)

Answer:

s:

• For problem 9: \( n = \frac{4}{5} \) or \( n = 1 \)
• For problem 10: \( r = -\frac{2}{3} \) or \( r = 1 \)