Question

1. $2x^2 - 3x - 15 = 5$
2. $x^2 + 2x - 1 = 2$
3. $2k^2 + 9k = -7$
4. $5r^2 = 80$
5. $2x^2 - 36 = x$
6. $5x^2 + 9x = -4$
7. $k^2 - 31 - 2k = -6 - 3k^2 - 2k$

Explanation:

Let's solve each quadratic equation one by one:

Problem 10: \( x^2 + 2x - 1 = 2 \)

Step 1: Rewrite in standard form

Subtract 2 from both sides: \( x^2 + 2x - 3 = 0 \)

Step 2: Factor the quadratic

Find two numbers that multiply to -3 and add to 2: 3 and -1. So, \( (x + 3)(x - 1) = 0 \)

Step 3: Solve for \( x \)

Set each factor equal to zero:
\( x + 3 = 0 \) gives \( x = -3 \)
\( x - 1 = 0 \) gives \( x = 1 \)

Step 1: Rewrite in standard form

Add 7 to both sides: \( 2k^2 + 9k + 7 = 0 \)

Step 2: Factor the quadratic

Find two numbers that multiply to \( 2 \times 7 = 14 \) and add to 9: 7 and 2. Rewrite the middle term:
\( 2k^2 + 2k + 7k + 7 = 0 \)
Factor by grouping:
\( 2k(k + 1) + 7(k + 1) = 0 \)
\( (2k + 7)(k + 1) = 0 \)

Step 3: Solve for \( k \)

\( 2k + 7 = 0 \) gives \( k = -\frac{7}{2} \)
\( k + 1 = 0 \) gives \( k = -1 \)

Step 1: Divide both sides by 5

\( r^2 = 16 \)

Step 2: Take square roots

\( r = \pm \sqrt{16} = \pm 4 \)

Answer:

\( x = -3 \) or \( x = 1 \)

Problem 11: \( 2k^2 + 9k = -7 \)