Find the condition that one root of `ax^(2)+bx+c=0` may be
(i) three tiems the other,
(ii) n times the other,
(iii) more than the other by h.

Answer

To solve the question, we need to find the conditions for the roots of the quadratic equation \( ax^2 + bx + c = 0 \) based on the relationships between the roots. We will break it down into three parts as given in the question.

### Part (i): One root is three times the other

1. **Let the roots be \( \alpha \) and \( 3\alpha \)**.
2. **Sum of the roots**: According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + 3\alpha = -\frac{b}{a} \]
This simplifies to:
\[ 4\alpha = -\frac{b}{a} \quad \Rightarrow \quad \alpha = -\frac{b}{4a} \]

3. **Substituting \( \alpha \) back into the equation**: Since \( \alpha \) and \( 3\alpha \) are roots, they must satisfy the equation \( ax^2 + bx + c = 0 \). Substituting \( x = \alpha \):
\[ a\alpha^2 + b\alpha + c = 0 \]
Substituting \( \alpha = -\frac{b}{4a} \):
\[ a\left(-\frac{b}{4a}\right)^2 + b\left(-\frac{b}{4a}\right) + c = 0 \]
Simplifying:
\[ a\left(\frac{b^2}{16a^2}\right) - \frac{b^2}{4a} + c = 0 \]
\[ \frac{b^2}{16a} - \frac{4b^2}{16a} + c = 0 \]
\[ -\frac{3b^2}{16a} + c = 0 \quad \Rightarrow \quad 16ac = 3b^2 \]

### Condition for Part (i):
\[ 16ac = 3b^2 \]

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### Part (ii): One root is \( n \) times the other

1. **Let the roots be \( \alpha \) and \( n\alpha \)**.
2. **Sum of the roots**:
\[ \alpha + n\alpha = -\frac{b}{a} \quad \Rightarrow \quad (1+n)\alpha = -\frac{b}{a} \quad \Rightarrow \quad \alpha = -\frac{b}{a(1+n)} \]

3. **Substituting \( \alpha \) back into the equation**:
\[ a\alpha^2 + b\alpha + c = 0 \]
Substituting \( \alpha = -\frac{b}{a(1+n)} \):
\[ a\left(-\frac{b}{a(1+n)}\right)^2 + b\left(-\frac{b}{a(1+n)}\right) + c = 0 \]
Simplifying:
\[ a\left(\frac{b^2}{a^2(1+n)^2}\right) - \frac{b^2}{a(1+n)} + c = 0 \]
\[ \frac{b^2}{a(1+n)^2} - \frac{b^2(1+n)}{a(1+n)^2} + c = 0 \]
\[ \frac{b^2 - b^2(1+n) + c(1+n)^2}{a(1+n)^2} = 0 \]
\[ c(1+n)^2 - b^2n = 0 \quad \Rightarrow \quad c(1+n)^2 = b^2n \]

### Condition for Part (ii):
\[ c(1+n)^2 = b^2n \]

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### Part (iii): One root is more than the other by \( h \)

1. **Let the roots be \( \alpha \) and \( \alpha + h \)**.
2. **Sum of the roots**:
\[ \alpha + (\alpha + h) = -\frac{b}{a} \quad \Rightarrow \quad 2\alpha + h = -\frac{b}{a} \quad \Rightarrow \quad 2\alpha = -\frac{b}{a} - h \quad \Rightarrow \quad \alpha = -\frac{b + ah}{2a} \]

3. **Substituting \( \alpha \) back into the equation**:
\[ a\alpha^2 + b\alpha + c = 0 \]
Substituting \( \alpha = -\frac{b + ah}{2a} \):
\[ a\left(-\frac{b + ah}{2a}\right)^2 + b\left(-\frac{b + ah}{2a}\right) + c = 0 \]
Simplifying:
\[ a\left(\frac{(b + ah)^2}{4a^2}\right) - \frac{b(b + ah)}{2a} + c = 0 \]
\[ \frac{(b + ah)^2}{4a} - \frac{b^2 + abh}{2a} + c = 0 \]
\[ \frac{(b + ah)^2 - 2(b^2 + abh)}{4a} + c = 0 \]
\[ (b^2 + 2abh + a^2h^2 - 2b^2 - 2abh + 4ac) = 0 \]
\[ -b^2 + a^2h^2 + 4ac = 0 \quad \Rightarrow \quad b^2 = a^2h^2 + 4ac \]

### Condition for Part (iii):
\[ b^2 = a^2h^2 + 4ac \]

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### Summary of Conditions:
1. For one root being three times the other: \( 16ac = 3b^2 \)
2. For one root being \( n \) times the other: \( c(1+n)^2 = b^2n \)
3. For one root being more than the other by \( h \): \( b^2 = a^2h^2 + 4ac \)