Answer :
Sure, let's solve each equation step by step to find the solutions.
### Equation 1: [tex]\( 25 - r^2 = 4r \)[/tex]
1. Rearrange the Equation:
Bring all terms to one side of the equation, so it is in standard quadratic form:
[tex]\[ 25 - r^2 - 4r = 0 \][/tex]
This simplifies to:
[tex]\[ -r^2 - 4r + 25 = 0 \][/tex]
To make it easier to work with, multiply through by -1:
[tex]\[ r^2 + 4r - 25 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( r^2 + 4r - 25 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -25 \)[/tex]:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-25)}}{2 \cdot 1} \][/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]
[tex]\[ r = -2 \pm \sqrt{29} \][/tex]
So the solutions for [tex]\( r \)[/tex] are:
[tex]\[ r = -2 + \sqrt{29} \quad \text{and} \quad r = -2 - \sqrt{29} \][/tex]
### Equation 2: [tex]\( 3x(x - 2) = -7 \)[/tex]
1. Rearrange the Equation:
First, expand the left-hand side and move all terms to one side:
[tex]\[ 3x^2 - 6x = -7 \][/tex]
[tex]\[ 3x^2 - 6x + 7 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( 3x^2 - 6x + 7 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 3 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 7 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 84}}{6} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{-48}}{6} \][/tex]
Since the discriminant is negative, we have complex solutions:
[tex]\[ x = \frac{6 \pm \sqrt{48} i}{6} \][/tex]
Simplifying further:
[tex]\[ x = \frac{6 \pm 2\sqrt{3} i}{6} \][/tex]
[tex]\[ x = 1 \pm \frac{2\sqrt{3} i}{3} \][/tex]
So the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = 1 - \frac{2\sqrt{3} i}{3} \quad \text{and} \quad x = 1 + \frac{2\sqrt{3} i}{3} \][/tex]
### Equation 3: [tex]\( (w - 8)(w + 5) = 14 \)[/tex]
1. Rearrange the Equation:
First, expand the left-hand side and move all terms to one side:
[tex]\[ w^2 - 8w + 5w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 54 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( w^2 - 3w - 54 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -54 \)[/tex]:
[tex]\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ w = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \][/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]
Therefore:
[tex]\[ w = \frac{3 + 15}{2} \quad \text{and} \quad w = \frac{3 - 15}{2} \][/tex]
[tex]\[ w = 9 \quad \text{and} \quad w = -6 \][/tex]
So the solutions for [tex]\( w \)[/tex] are:
[tex]\[ w = 9 \quad \text{and} \quad w = -6 \][/tex]
To summarize, the solutions for the given equations are:
1. [tex]\( r = -2 + \sqrt{29} \)[/tex] and [tex]\( r = -2 - \sqrt{29} \)[/tex]
2. [tex]\( x = 1 - \frac{2\sqrt{3} i}{3} \)[/tex] and [tex]\( x = 1 + \frac{2\sqrt{3} i}{3} \)[/tex]
3. [tex]\( w = 9 \)[/tex] and [tex]\( w = -6 \)[/tex]
### Equation 1: [tex]\( 25 - r^2 = 4r \)[/tex]
1. Rearrange the Equation:
Bring all terms to one side of the equation, so it is in standard quadratic form:
[tex]\[ 25 - r^2 - 4r = 0 \][/tex]
This simplifies to:
[tex]\[ -r^2 - 4r + 25 = 0 \][/tex]
To make it easier to work with, multiply through by -1:
[tex]\[ r^2 + 4r - 25 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( r^2 + 4r - 25 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = -25 \)[/tex]:
[tex]\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ r = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-25)}}{2 \cdot 1} \][/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]
[tex]\[ r = -2 \pm \sqrt{29} \][/tex]
So the solutions for [tex]\( r \)[/tex] are:
[tex]\[ r = -2 + \sqrt{29} \quad \text{and} \quad r = -2 - \sqrt{29} \][/tex]
### Equation 2: [tex]\( 3x(x - 2) = -7 \)[/tex]
1. Rearrange the Equation:
First, expand the left-hand side and move all terms to one side:
[tex]\[ 3x^2 - 6x = -7 \][/tex]
[tex]\[ 3x^2 - 6x + 7 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( 3x^2 - 6x + 7 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 3 \)[/tex], [tex]\( b = -6 \)[/tex], and [tex]\( c = 7 \)[/tex]:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot 7}}{2 \cdot 3} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{36 - 84}}{6} \][/tex]
[tex]\[ x = \frac{6 \pm \sqrt{-48}}{6} \][/tex]
Since the discriminant is negative, we have complex solutions:
[tex]\[ x = \frac{6 \pm \sqrt{48} i}{6} \][/tex]
Simplifying further:
[tex]\[ x = \frac{6 \pm 2\sqrt{3} i}{6} \][/tex]
[tex]\[ x = 1 \pm \frac{2\sqrt{3} i}{3} \][/tex]
So the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = 1 - \frac{2\sqrt{3} i}{3} \quad \text{and} \quad x = 1 + \frac{2\sqrt{3} i}{3} \][/tex]
### Equation 3: [tex]\( (w - 8)(w + 5) = 14 \)[/tex]
1. Rearrange the Equation:
First, expand the left-hand side and move all terms to one side:
[tex]\[ w^2 - 8w + 5w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 40 = 14 \][/tex]
[tex]\[ w^2 - 3w - 54 = 0 \][/tex]
2. Solve the Quadratic Equation:
To solve [tex]\( w^2 - 3w - 54 = 0 \)[/tex], use the quadratic formula where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -54 \)[/tex]:
[tex]\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Substituting the values in:
[tex]\[ w = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-54)}}{2 \cdot 1} \][/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]
Therefore:
[tex]\[ w = \frac{3 + 15}{2} \quad \text{and} \quad w = \frac{3 - 15}{2} \][/tex]
[tex]\[ w = 9 \quad \text{and} \quad w = -6 \][/tex]
So the solutions for [tex]\( w \)[/tex] are:
[tex]\[ w = 9 \quad \text{and} \quad w = -6 \][/tex]
To summarize, the solutions for the given equations are:
1. [tex]\( r = -2 + \sqrt{29} \)[/tex] and [tex]\( r = -2 - \sqrt{29} \)[/tex]
2. [tex]\( x = 1 - \frac{2\sqrt{3} i}{3} \)[/tex] and [tex]\( x = 1 + \frac{2\sqrt{3} i}{3} \)[/tex]
3. [tex]\( w = 9 \)[/tex] and [tex]\( w = -6 \)[/tex]
