Answer :
To solve for \( s \) in the equation \( x = \sqrt{\frac{s(s-a)}{bc}} \), follow these steps:
1. Square both sides: To eliminate the square root, square both sides of the equation:
[tex]\[ x^2 = \left( \sqrt{\frac{s(s-a)}{bc}} \right)^2 \][/tex]
2. Simplify: The right side simplifies to remove the square root:
[tex]\[ x^2 = \frac{s(s-a)}{bc} \][/tex]
3. Isolate \( s(s-a) \): Multiply both sides of the equation by \( bc \) to remove the fraction:
[tex]\[ x^2 \cdot bc = s(s-a) \][/tex]
4. Distribute: Distribute \( s \) on the right side:
[tex]\[ bcx^2 = s^2 - as \][/tex]
5. Rearrange into a standard quadratic form: Bring all terms to one side of the equation:
[tex]\[ s^2 - as - bcx^2 = 0 \][/tex]
6. Use the quadratic formula: The standard form of a quadratic equation is \( Ax^2 + Bx + C = 0 \). Here, \( A = 1 \), \( B = -a \), and \( C = -bcx^2 \). The quadratic formula \( s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) can be used to solve for \( s \):
[tex]\[ s = \frac{-(-a) \pm \sqrt{(-a)^2 - 4 \cdot 1 \cdot (-bcx^2)}}{2 \cdot 1} \][/tex]
7. Simplify under the square root and solve:
[tex]\[ s = \frac{a \pm \sqrt{a^2 + 4bcx^2}}{2} \][/tex]
Therefore, the solutions for \( s \) are:
[tex]\[ s = \frac{a + \sqrt{a^2 + 4bcx^2}}{2} \quad \text{and} \quad s = \frac{a - \sqrt{a^2 + 4bcx^2}}{2} \][/tex]
1. Square both sides: To eliminate the square root, square both sides of the equation:
[tex]\[ x^2 = \left( \sqrt{\frac{s(s-a)}{bc}} \right)^2 \][/tex]
2. Simplify: The right side simplifies to remove the square root:
[tex]\[ x^2 = \frac{s(s-a)}{bc} \][/tex]
3. Isolate \( s(s-a) \): Multiply both sides of the equation by \( bc \) to remove the fraction:
[tex]\[ x^2 \cdot bc = s(s-a) \][/tex]
4. Distribute: Distribute \( s \) on the right side:
[tex]\[ bcx^2 = s^2 - as \][/tex]
5. Rearrange into a standard quadratic form: Bring all terms to one side of the equation:
[tex]\[ s^2 - as - bcx^2 = 0 \][/tex]
6. Use the quadratic formula: The standard form of a quadratic equation is \( Ax^2 + Bx + C = 0 \). Here, \( A = 1 \), \( B = -a \), and \( C = -bcx^2 \). The quadratic formula \( s = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \) can be used to solve for \( s \):
[tex]\[ s = \frac{-(-a) \pm \sqrt{(-a)^2 - 4 \cdot 1 \cdot (-bcx^2)}}{2 \cdot 1} \][/tex]
7. Simplify under the square root and solve:
[tex]\[ s = \frac{a \pm \sqrt{a^2 + 4bcx^2}}{2} \][/tex]
Therefore, the solutions for \( s \) are:
[tex]\[ s = \frac{a + \sqrt{a^2 + 4bcx^2}}{2} \quad \text{and} \quad s = \frac{a - \sqrt{a^2 + 4bcx^2}}{2} \][/tex]
