Answer :
Certainly! Let's solve each of the given equations step-by-step.
### Equation 1: [tex]\( 9(a + 7) = 5a^2 - 3 \)[/tex]
1. Distribute the 9 on the left-hand side:
[tex]\[ 9a + 63 = 5a^2 - 3 \][/tex]
2. Move all terms to one side to set the equation to 0:
[tex]\[ 5a^2 - 9a - 66 = 0 \][/tex]
3. Use the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 5 \)[/tex], [tex]\( b = -9 \)[/tex], and [tex]\( c = -66 \)[/tex]:
[tex]\[ a = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 5 \cdot (-66)}}{10} \][/tex]
4. Simplify the expression inside the square root:
[tex]\[ a = \frac{9 \pm \sqrt{81 + 1320}}{10} \][/tex]
[tex]\[ a = \frac{9 \pm \sqrt{1401}}{10} \][/tex]
5. Finally, the solutions are:
[tex]\[ a = \frac{9 - \sqrt{1401}}{10} \quad \text{and} \quad a = \frac{9 + \sqrt{1401}}{10} \][/tex]
### Equation 2: [tex]\( 2k^2 = 3 - 5k \)[/tex]
1. Move all terms to one side:
[tex]\[ 2k^2 + 5k - 3 = 0 \][/tex]
2. Again, use the quadratic formula where [tex]\( a = 2 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \][/tex]
3. Simplify the expression inside the square root:
[tex]\[ k = \frac{-5 \pm \sqrt{25 + 24}}{4} \][/tex]
[tex]\[ k = \frac{-5 \pm \sqrt{49}}{4} \][/tex]
4. Simplify further:
[tex]\[ k = \frac{-5 \pm 7}{4} \][/tex]
5. The solutions are:
[tex]\[ k = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad k = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \][/tex]
### Equation 3: [tex]\( 3(a - 5)^2 = 2a \)[/tex]
1. Expand the equation:
[tex]\[ 3(a^2 - 10a + 25) = 2a \][/tex]
[tex]\[ 3a^2 - 30a + 75 = 2a \][/tex]
2. Move all terms to one side:
[tex]\[ 3a^2 - 32a + 75 = 0 \][/tex]
3. Use the quadratic formula where [tex]\( a = 3 \)[/tex], [tex]\( b = -32 \)[/tex], and [tex]\( c = 75 \)[/tex]:
[tex]\[ a = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 3 \cdot 75}}{2 \cdot 3} \][/tex]
4. Simplify the expression inside the square root:
[tex]\[ a = \frac{32 \pm \sqrt{1024 - 900}}{6} \][/tex]
[tex]\[ a = \frac{32 \pm \sqrt{124}}{6} \][/tex]
5. Simplify further:
[tex]\[ a = \frac{32 \pm 2\sqrt{31}}{6} \][/tex]
[tex]\[ a = \frac{16 \pm \sqrt{31}}{3} \][/tex]
6. The solutions are:
[tex]\[ a = \frac{16 - \sqrt{31}}{3} \quad \text{and} \quad a = \frac{16 + \sqrt{31}}{3} \][/tex]
### Summary of Solutions:
- Equation 1 solutions:
[tex]\[ a = \frac{9 - \sqrt{1401}}{10} \quad \text{and} \quad a = \frac{9 + \sqrt{1401}}{10} \][/tex]
- Equation 2 solutions:
[tex]\[ k = -3 \quad \text{and} \quad k = \frac{1}{2} \][/tex]
- Equation 3 solutions:
[tex]\[ a = \frac{16 - \sqrt{31}}{3} \quad \text{and} \quad a = \frac{16 + \sqrt{31}}{3} \][/tex]
### Equation 1: [tex]\( 9(a + 7) = 5a^2 - 3 \)[/tex]
1. Distribute the 9 on the left-hand side:
[tex]\[ 9a + 63 = 5a^2 - 3 \][/tex]
2. Move all terms to one side to set the equation to 0:
[tex]\[ 5a^2 - 9a - 66 = 0 \][/tex]
3. Use the quadratic formula [tex]\( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex] where [tex]\( a = 5 \)[/tex], [tex]\( b = -9 \)[/tex], and [tex]\( c = -66 \)[/tex]:
[tex]\[ a = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 5 \cdot (-66)}}{10} \][/tex]
4. Simplify the expression inside the square root:
[tex]\[ a = \frac{9 \pm \sqrt{81 + 1320}}{10} \][/tex]
[tex]\[ a = \frac{9 \pm \sqrt{1401}}{10} \][/tex]
5. Finally, the solutions are:
[tex]\[ a = \frac{9 - \sqrt{1401}}{10} \quad \text{and} \quad a = \frac{9 + \sqrt{1401}}{10} \][/tex]
### Equation 2: [tex]\( 2k^2 = 3 - 5k \)[/tex]
1. Move all terms to one side:
[tex]\[ 2k^2 + 5k - 3 = 0 \][/tex]
2. Again, use the quadratic formula where [tex]\( a = 2 \)[/tex], [tex]\( b = 5 \)[/tex], and [tex]\( c = -3 \)[/tex]:
[tex]\[ k = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 2 \cdot (-3)}}{2 \cdot 2} \][/tex]
3. Simplify the expression inside the square root:
[tex]\[ k = \frac{-5 \pm \sqrt{25 + 24}}{4} \][/tex]
[tex]\[ k = \frac{-5 \pm \sqrt{49}}{4} \][/tex]
4. Simplify further:
[tex]\[ k = \frac{-5 \pm 7}{4} \][/tex]
5. The solutions are:
[tex]\[ k = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{and} \quad k = \frac{-5 - 7}{4} = \frac{-12}{4} = -3 \][/tex]
### Equation 3: [tex]\( 3(a - 5)^2 = 2a \)[/tex]
1. Expand the equation:
[tex]\[ 3(a^2 - 10a + 25) = 2a \][/tex]
[tex]\[ 3a^2 - 30a + 75 = 2a \][/tex]
2. Move all terms to one side:
[tex]\[ 3a^2 - 32a + 75 = 0 \][/tex]
3. Use the quadratic formula where [tex]\( a = 3 \)[/tex], [tex]\( b = -32 \)[/tex], and [tex]\( c = 75 \)[/tex]:
[tex]\[ a = \frac{32 \pm \sqrt{(-32)^2 - 4 \cdot 3 \cdot 75}}{2 \cdot 3} \][/tex]
4. Simplify the expression inside the square root:
[tex]\[ a = \frac{32 \pm \sqrt{1024 - 900}}{6} \][/tex]
[tex]\[ a = \frac{32 \pm \sqrt{124}}{6} \][/tex]
5. Simplify further:
[tex]\[ a = \frac{32 \pm 2\sqrt{31}}{6} \][/tex]
[tex]\[ a = \frac{16 \pm \sqrt{31}}{3} \][/tex]
6. The solutions are:
[tex]\[ a = \frac{16 - \sqrt{31}}{3} \quad \text{and} \quad a = \frac{16 + \sqrt{31}}{3} \][/tex]
### Summary of Solutions:
- Equation 1 solutions:
[tex]\[ a = \frac{9 - \sqrt{1401}}{10} \quad \text{and} \quad a = \frac{9 + \sqrt{1401}}{10} \][/tex]
- Equation 2 solutions:
[tex]\[ k = -3 \quad \text{and} \quad k = \frac{1}{2} \][/tex]
- Equation 3 solutions:
[tex]\[ a = \frac{16 - \sqrt{31}}{3} \quad \text{and} \quad a = \frac{16 + \sqrt{31}}{3} \][/tex]
