Sure, let's solve each equation step by step.
### Problem 6
Solve the equation: [tex]\( 25 - r^2 = 4r \)[/tex]
1. Rearrange the equation to set it equal to zero:
[tex]\[ 25 - r^2 - 4r = 0 \][/tex]
2. This is a quadratic equation in the form [tex]\( ar^2 + br + c = 0 \)[/tex]:
[tex]\[ -r^2 - 4r + 25 = 0 \][/tex]
3. Use the quadratic formula [tex]\( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = 25 \)[/tex]:
[tex]\[ r = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(-1)(25)}}{2(-1)} \][/tex]
[tex]\[ r = \frac{4 \pm \sqrt{16 + 100}}{-2} \][/tex]
[tex]\[ r = \frac{4 \pm \sqrt{116}}{-2} \][/tex]
[tex]\[ r = \frac{4 \pm 2\sqrt{29}}{-2} \][/tex]
4. Simplify:
[tex]\[ r = -2 \pm \sqrt{29} \][/tex]
So, the solutions are:
[tex]\[ r = -2 + \sqrt{29} \quad \text{and} \quad r = -2 - \sqrt{29} \][/tex]
### Problem 7
Solve the equation: [tex]\( 3x(x-2) = -7 \)[/tex]
1. Distribute and rearrange the equation to set it equal to zero:
[tex]\[ 3x^2 - 6x + 7 = 0 \][/tex]
2. This is a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ 3x^2 - 6x + 7 = 0 \][/tex]
3. Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 3 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 7 \)[/tex]:
[tex]\[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(3)(7)}}{2(3)} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 84}}{6} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{-48}}{6} \][/tex]
[tex]\[ x = \frac{6 \pm 2i\sqrt{12}}{6} \][/tex]
[tex]\[ x = 1 \pm \frac{2i\sqrt{3}}{3} \][/tex]
So, the solutions are:
[tex]\[ x = 1 - \frac{2i\sqrt{3}}{3} \quad \text{and} \quad x = 1 + \frac{2i\sqrt{3}}{3} \][/tex]
### Problem 8
Solve the equation: [tex]\( \frac{1}{2}(h-6) = 0 \)[/tex]
1. Multiply both sides by 2 to eliminate the fraction:
[tex]\[ h - 6 = 0 \][/tex]
2. Solve for [tex]\( h \)[/tex]:
[tex]\[ h = 6 \][/tex]
So, the solution is:
[tex]\[ h = 6 \][/tex]
### Problem 9
Solve the equation: [tex]\( (x+2)^2 = 0 \)[/tex]
1. Take the square root of both sides:
[tex]\[ x+2 = 0 \][/tex]
2. Solve for [tex]\( x \)[/tex]:
[tex]\[ x = -2 \][/tex]
So, the solution is:
[tex]\[ x = -2 \][/tex]
### Problem 10
Solve the equation: [tex]\( (w-8)(w+5) = 14 \)[/tex]
1. Expand the left side and rearrange the equation to set it equal to zero:
[tex]\[ w^2 - 8w + 5w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 54 = 0 \][/tex]
2. This is a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex]:
[tex]\[ w^2 - 3w - 54 = 0 \][/tex]
3. Use the quadratic formula [tex]\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -54 \)[/tex]:
[tex]\[ w = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-54)}}{2(1)} \][/tex]
[tex]\[ w = \frac{3 \pm \sqrt{9 + 216}}{2} \][/tex]
[tex]\[ w = \frac{3 \pm \sqrt{225}}{2} \][/tex]
[tex]\[ w = \frac{3 \pm 15}{2} \][/tex]
4. Simplify:
[tex]\[ w = \frac{3 + 15}{2} = 9 \][/tex]
[tex]\[ w = \frac{3 - 15}{2} = -6 \][/tex]
So, the solutions are:
[tex]\[ w = 9 \quad \text{and} \quad w = -6 \][/tex]
