Answer :
Sure! Let's solve each of these equations step-by-step:
### Problem 1: [tex]\(3x^2 - 7 = 2x\)[/tex]
1. Move all terms to one side of the equation:
[tex]\[ 3x^2 - 2x - 7 = 0 \][/tex]
2. Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -7 \)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-7)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 84}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{88}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm 2\sqrt{22}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{22}}{3} \][/tex]
3. So, the solutions are:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{22}}{3} \quad \text{and} \quad x = \frac{1}{3} + \frac{\sqrt{22}}{3} \][/tex]
### Problem 2: [tex]\(15x + x^2 - 2 = 0\)[/tex]
1. This is a standard form of a quadratic equation [tex]\( x^2 + 15x - 2 = 0 \)[/tex].
2. Use the quadratic formula with [tex]\( a = 1 \)[/tex], [tex]\( b = 15 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-15 \pm \sqrt{225 + 8}}{2} \][/tex]
[tex]\[ x = \frac{-15 \pm \sqrt{233}}{2} \][/tex]
3. So, the solutions are:
[tex]\[ x = \frac{-15}{2} + \frac{\sqrt{233}}{2} \quad \text{and} \quad x = \frac{-15}{2} - \frac{\sqrt{233}}{2} \][/tex]
### Problem 3: [tex]\(-10(x - 5) = 12\)[/tex]
1. Distribute the [tex]\(-10\)[/tex]:
[tex]\[ -10x + 50 = 12 \][/tex]
2. Move the constant term to the other side:
[tex]\[ -10x = 12 - 50 \][/tex]
[tex]\[ -10x = -38 \][/tex]
3. Divide by [tex]\(-10\)[/tex]:
[tex]\[ x = \frac{-38}{-10} \][/tex]
[tex]\[ x = \frac{19}{5} \][/tex]
### Problem 4: [tex]\(16(2m^2 + 4) = 0\)[/tex]
1. Notice that the equation is set to zero:
[tex]\[ 16(2m^2 + 4) = 0 \][/tex]
2. Divide both sides by 16:
[tex]\[ 2m^2 + 4 = 0 \][/tex]
3. Solve for [tex]\( m^2 \)[/tex]:
[tex]\[ 2m^2 = -4 \][/tex]
[tex]\[ m^2 = -2 \][/tex]
4. Since [tex]\( m^2 = -2 \)[/tex] involves a negative under the square, the solutions are imaginary:
[tex]\[ m = \pm \sqrt{-2} \][/tex]
[tex]\[ m = \pm i\sqrt{2} \][/tex]
5. So, the solutions are:
[tex]\[ m = \sqrt{2}i \quad \text{and} \quad m = -\sqrt{2}i \][/tex]
### Problem 5: [tex]\((x + 3)(x + 11) = 0\)[/tex]
1. Set each factor equal to zero:
[tex]\[ x + 3 = 0 \quad \text{or} \quad x + 11 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -3 \quad \text{or} \quad x = -11 \][/tex]
To summarize, the solutions are:
1. [tex]\( x = \frac{1}{3} - \frac{\sqrt{22}}{3}, \frac{1}{3} + \frac{\sqrt{22}}{3} \)[/tex]
2. [tex]\( x = \frac{-15}{2} + \frac{\sqrt{233}}{2}, \frac{-15}{2} - \frac{\sqrt{233}}{2} \)[/tex]
3. [tex]\( x = \frac{19}{5} \)[/tex]
4. [tex]\( m = \sqrt{2}i, -\sqrt{2}i \)[/tex]
5. [tex]\( x = -3, -11 \)[/tex]
### Problem 1: [tex]\(3x^2 - 7 = 2x\)[/tex]
1. Move all terms to one side of the equation:
[tex]\[ 3x^2 - 2x - 7 = 0 \][/tex]
2. Solve this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -7 \)[/tex]:
[tex]\[ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 3 \cdot (-7)}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{4 + 84}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm \sqrt{88}}{6} \][/tex]
[tex]\[ x = \frac{2 \pm 2\sqrt{22}}{6} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{22}}{3} \][/tex]
3. So, the solutions are:
[tex]\[ x = \frac{1}{3} - \frac{\sqrt{22}}{3} \quad \text{and} \quad x = \frac{1}{3} + \frac{\sqrt{22}}{3} \][/tex]
### Problem 2: [tex]\(15x + x^2 - 2 = 0\)[/tex]
1. This is a standard form of a quadratic equation [tex]\( x^2 + 15x - 2 = 0 \)[/tex].
2. Use the quadratic formula with [tex]\( a = 1 \)[/tex], [tex]\( b = 15 \)[/tex], and [tex]\( c = -2 \)[/tex]:
[tex]\[ x = \frac{-15 \pm \sqrt{15^2 - 4 \cdot 1 \cdot (-2)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-15 \pm \sqrt{225 + 8}}{2} \][/tex]
[tex]\[ x = \frac{-15 \pm \sqrt{233}}{2} \][/tex]
3. So, the solutions are:
[tex]\[ x = \frac{-15}{2} + \frac{\sqrt{233}}{2} \quad \text{and} \quad x = \frac{-15}{2} - \frac{\sqrt{233}}{2} \][/tex]
### Problem 3: [tex]\(-10(x - 5) = 12\)[/tex]
1. Distribute the [tex]\(-10\)[/tex]:
[tex]\[ -10x + 50 = 12 \][/tex]
2. Move the constant term to the other side:
[tex]\[ -10x = 12 - 50 \][/tex]
[tex]\[ -10x = -38 \][/tex]
3. Divide by [tex]\(-10\)[/tex]:
[tex]\[ x = \frac{-38}{-10} \][/tex]
[tex]\[ x = \frac{19}{5} \][/tex]
### Problem 4: [tex]\(16(2m^2 + 4) = 0\)[/tex]
1. Notice that the equation is set to zero:
[tex]\[ 16(2m^2 + 4) = 0 \][/tex]
2. Divide both sides by 16:
[tex]\[ 2m^2 + 4 = 0 \][/tex]
3. Solve for [tex]\( m^2 \)[/tex]:
[tex]\[ 2m^2 = -4 \][/tex]
[tex]\[ m^2 = -2 \][/tex]
4. Since [tex]\( m^2 = -2 \)[/tex] involves a negative under the square, the solutions are imaginary:
[tex]\[ m = \pm \sqrt{-2} \][/tex]
[tex]\[ m = \pm i\sqrt{2} \][/tex]
5. So, the solutions are:
[tex]\[ m = \sqrt{2}i \quad \text{and} \quad m = -\sqrt{2}i \][/tex]
### Problem 5: [tex]\((x + 3)(x + 11) = 0\)[/tex]
1. Set each factor equal to zero:
[tex]\[ x + 3 = 0 \quad \text{or} \quad x + 11 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -3 \quad \text{or} \quad x = -11 \][/tex]
To summarize, the solutions are:
1. [tex]\( x = \frac{1}{3} - \frac{\sqrt{22}}{3}, \frac{1}{3} + \frac{\sqrt{22}}{3} \)[/tex]
2. [tex]\( x = \frac{-15}{2} + \frac{\sqrt{233}}{2}, \frac{-15}{2} - \frac{\sqrt{233}}{2} \)[/tex]
3. [tex]\( x = \frac{19}{5} \)[/tex]
4. [tex]\( m = \sqrt{2}i, -\sqrt{2}i \)[/tex]
5. [tex]\( x = -3, -11 \)[/tex]
