Question

1. solve the equation for all real solutions in simplest form.\\( k^2 - 2k - 11 = 3k \\)\\( \\)\\( \\)13. solve the equation for all real solutions in simplest form.\\( 6k^2 + 3k - 4 = 2k^2 \\)\\( \\)\\( \\)16. solve the equation for all real solutions in simplest form.\\( 3t^2 + 12t + 14 = 4 \\)\\( \\)\\( \\)17. solve the equation for all real solutions in simplest form.\\( 4d^2 + 4d + 1 = -6d \\)\\( \\)\\( \\)18. solve the equation for all real solutions in simplest form.\\( 3k^2 + 2k - 3 = 0 \\)

Explanation:

Problem 12: Solve \( k^2 - 2k - 11 = 3k \)

Step 1: Rearrange to standard quadratic form

Subtract \( 3k \) from both sides: \( k^2 - 5k - 11 = 0 \)

Step 2: Apply quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 1 \), \( b = -5 \), \( c = -11 \).
Discriminant: \( (-5)^2 - 4(1)(-11) = 25 + 44 = 69 \)
Solutions: \( k = \frac{5 \pm \sqrt{69}}{2} \)

Step 1: Rearrange to standard quadratic form

Subtract \( 2k^2 \) from both sides: \( 4k^2 + 3k - 4 = 0 \)

Step 2: Apply quadratic formula \( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 4 \), \( b = 3 \), \( c = -4 \).
Discriminant: \( 3^2 - 4(4)(-4) = 9 + 64 = 73 \)
Solutions: \( k = \frac{-3 \pm \sqrt{73}}{8} \)

Step 1: Rearrange to standard quadratic form

Subtract \( 4 \) from both sides: \( 3t^2 + 12t + 10 = 0 \)

Step 2: Apply quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Here, \( a = 3 \), \( b = 12 \), \( c = 10 \).
Discriminant: \( 12^2 - 4(3)(10) = 144 - 120 = 24 \)
Simplify \( \sqrt{24} = 2\sqrt{6} \), so solutions: \( t = \frac{-12 \pm 2\sqrt{6}}{6} = \frac{-6 \pm \sqrt{6}}{3} \)

Answer:

\( k = \frac{5 + \sqrt{69}}{2} \) or \( k = \frac{5 - \sqrt{69}}{2} \)

Problem 13: Solve \( 6k^2 + 3k - 4 = 2k^2 \)