If `alpha, beta` are the roots of the quadratic equation `cx^2- 2bx + 4a = 0` then find the quadratic equation whose roots are:
(i) `alpha/2, beta/2`, (ii) `alpha^(2), beta^(2)`, (iii) `alpha + 1, beta+1`, (iv) `(1+alpha)/(1-alpha) (1+beta)/(1-beta)`, (v) `alpha/beta, beta/alpha`

Answer

To find the quadratic equations whose roots are derived from the roots \( \alpha \) and \( \beta \) of the quadratic equation \( cx^2 - 2bx + 4a = 0 \), we will use the relationships between the roots and coefficients of the quadratic equation.

### Given:
The roots \( \alpha \) and \( \beta \) satisfy:
1. Sum of the roots: \( \alpha + \beta = \frac{2b}{c} \)
2. Product of the roots: \( \alpha \beta = \frac{4a}{c} \)

### (i) Roots: \( \frac{\alpha}{2}, \frac{\beta}{2} \)

**Step 1:** Calculate the sum of the new roots:
\[ \frac{\alpha}{2} + \frac{\beta}{2} = \frac{\alpha + \beta}{2} = \frac{2b}{2c} = \frac{b}{c} \]

**Step 2:** Calculate the product of the new roots:
\[ \frac{\alpha}{2} \cdot \frac{\beta}{2} = \frac{\alpha \beta}{4} = \frac{4a}{4c} = \frac{a}{c} \]

**Step 3:** Form the quadratic equation:
\[ x^2 - \left(\frac{b}{c}\right)x + \frac{a}{c} = 0 \]
Multiplying through by \( c \):
\[ cx^2 - bx + a = 0 \]

### (ii) Roots: \( \alpha^2, \beta^2 \)

**Step 1:** Calculate the sum of the new roots:
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(\frac{2b}{c}\right)^2 - 2\left(\frac{4a}{c}\right) = \frac{4b^2}{c^2} - \frac{8a}{c} \]

**Step 2:** Calculate the product of the new roots:
\[ \alpha^2 \beta^2 = (\alpha \beta)^2 = \left(\frac{4a}{c}\right)^2 = \frac{16a^2}{c^2} \]

**Step 3:** Form the quadratic equation:
\[ x^2 - \left(\frac{4b^2 - 8ac}{c^2}\right)x + \frac{16a^2}{c^2} = 0 \]
Multiplying through by \( c^2 \):
\[ c^2x^2 - (4b^2 - 8ac)x + 16a^2 = 0 \]

### (iii) Roots: \( \alpha + 1, \beta + 1 \)

**Step 1:** Calculate the sum of the new roots:
\[ (\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{2b}{c} + 2 = \frac{2b + 2c}{c} = \frac{2(b + c)}{c} \]

**Step 2:** Calculate the product of the new roots:
\[ (\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = \frac{4a}{c} + \frac{2b}{c} + 1 = \frac{4a + 2b + c}{c} \]

**Step 3:** Form the quadratic equation:
\[ x^2 - \left(\frac{2(b + c)}{c}\right)x + \frac{4a + 2b + c}{c} = 0 \]
Multiplying through by \( c \):
\[ cx^2 - 2(b + c)x + (4a + 2b + c) = 0 \]

### (iv) Roots: \( \frac{1 + \alpha}{1 - \alpha}, \frac{1 + \beta}{1 - \beta} \)

**Step 1:** Calculate the sum of the new roots:
\[ \frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} = \frac{(1 + \alpha)(1 - \beta) + (1 + \beta)(1 - \alpha)}{(1 - \alpha)(1 - \beta)} \]
After simplifying, we find:
\[ = \frac{2 + \alpha + \beta - (\alpha\beta + \alpha + \beta)}{1 - (\alpha + \beta) + \alpha\beta} = \frac{2 - \alpha\beta}{1 - \frac{2b}{c} + \frac{4a}{c}} = \frac{2 - \frac{4a}{c}}{\frac{c - 2b + 4a}{c}} = \frac{2c - 4a}{c - 2b + 4a} \]

**Step 2:** Calculate the product of the new roots:
\[ \frac{(1 + \alpha)(1 + \beta)}{(1 - \alpha)(1 - \beta)} = \frac{1 + \alpha + \beta + \alpha\beta}{1 - (\alpha + \beta) + \alpha\beta} = \frac{1 + \frac{2b}{c} + \frac{4a}{c}}{1 - \frac{2b}{c} + \frac{4a}{c}} = \frac{c + 2b + 4a}{c - 2b + 4a} \]

**Step 3:** Form the quadratic equation:
\[ x^2 - \left(\frac{2c - 4a}{c - 2b + 4a}\right)x + \frac{c + 2b + 4a}{c - 2b + 4a} = 0 \]
Multiplying through by \( (c - 2b + 4a) \):
\[ (c - 2b + 4a)x^2 - (2c - 4a)x + (c + 2b + 4a) = 0 \]

### (v) Roots: \( \frac{\alpha}{\beta}, \frac{\beta}{\alpha} \)

**Step 1:** Calculate the sum of the new roots:
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \frac{\left(\frac{2b}{c}\right)^2 - 2\left(\frac{4a}{c}\right)}{\frac{4a}{c}} = \frac{\frac{4b^2}{c^2} - \frac{8a}{c}}{\frac{4a}{c}} = \frac{4b^2 - 8ac}{4ac} = \frac{b^2 - 2ac}{ac} \]

**Step 2:** Calculate the product of the new roots:
\[ \frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1 \]

**Step 3:** Form the quadratic equation:
\[ x^2 - \left(\frac{b^2 - 2ac}{ac}\right)x + 1 = 0 \]
Multiplying through by \( ac \):
\[ acx^2 - (b^2 - 2ac)x + ac = 0 \]

### Summary of Quadratic Equations:
1. \( cx^2 - bx + a = 0 \)
2. \( c^2x^2 - (4b^2 - 8ac)x + 16a^2 = 0 \)
3. \( cx^2 - 2(b + c)x + (4a + 2b + c) = 0 \)
4. \( (c - 2b + 4a)x^2 - (2c - 4a)x + (c + 2b + 4a) = 0 \)
5. \( acx^2 - (b^2 - 2ac)x + ac = 0 \)