Find the roots of the following quadratic equations by the method of completing the square :
(i) `x^(2)-10-24=0` (ii) `2x^(2)-7x-39=0`
(iii) `5x^(2)+6x-8=0`

Answer

To solve the quadratic equations by the method of completing the square, we will follow these steps for each equation.

### (i) Solve \( x^2 - 10x - 24 = 0 \)

1. **Rearrange the equation**:
\[ x^2 - 10x = 24 \]

2. **Complete the square**:
- Take half of the coefficient of \( x \) (which is \(-10\)), square it, and add it to both sides.
- Half of \(-10\) is \(-5\), and squaring it gives \(25\).
\[ x^2 - 10x + 25 = 24 + 25 \]

3. **Rewrite the left side as a square**:
\[ (x - 5)^2 = 49 \]

4. **Take the square root of both sides**:
\[ x - 5 = \pm 7 \]

5. **Solve for \( x \)**:
- For \( x - 5 = 7 \):
\[ x = 12 \]
- For \( x - 5 = -7 \):
\[ x = -2 \]

**Roots**: \( x = 12 \) and \( x = -2 \)

### (ii) Solve \( 2x^2 - 7x - 39 = 0 \)

1. **Rearrange the equation**:
\[ 2x^2 - 7x = 39 \]

2. **Divide the entire equation by 2** to simplify:
\[ x^2 - \frac{7}{2}x = \frac{39}{2} \]

3. **Complete the square**:
- Half of \(-\frac{7}{2}\) is \(-\frac{7}{4}\), and squaring it gives \(\frac{49}{16}\).
\[ x^2 - \frac{7}{2}x + \frac{49}{16} = \frac{39}{2} + \frac{49}{16} \]

4. **Rewrite the left side as a square**:
\[ \left(x - \frac{7}{4}\right)^2 = \frac{312}{16} + \frac{49}{16} = \frac{361}{16} \]

5. **Take the square root of both sides**:
\[ x - \frac{7}{4} = \pm \frac{19}{4} \]

6. **Solve for \( x \)**:
- For \( x - \frac{7}{4} = \frac{19}{4} \):
\[ x = \frac{26}{4} = \frac{13}{2} \]
- For \( x - \frac{7}{4} = -\frac{19}{4} \):
\[ x = -\frac{12}{4} = -3 \]

**Roots**: \( x = \frac{13}{2} \) and \( x = -3 \)

### (iii) Solve \( 5x^2 + 6x - 8 = 0 \)

1. **Rearrange the equation**:
\[ 5x^2 + 6x = 8 \]

2. **Divide the entire equation by 5** to simplify:
\[ x^2 + \frac{6}{5}x = \frac{8}{5} \]

3. **Complete the square**:
- Half of \(\frac{6}{5}\) is \(\frac{3}{5}\), and squaring it gives \(\frac{9}{25}\).
\[ x^2 + \frac{6}{5}x + \frac{9}{25} = \frac{8}{5} + \frac{9}{25} \]

4. **Rewrite the left side as a square**:
\[ \left(x + \frac{3}{5}\right)^2 = \frac{40}{25} + \frac{9}{25} = \frac{49}{25} \]

5. **Take the square root of both sides**:
\[ x + \frac{3}{5} = \pm \frac{7}{5} \]

6. **Solve for \( x \)**:
- For \( x + \frac{3}{5} = \frac{7}{5} \):
\[ x = \frac{4}{5} \]
- For \( x + \frac{3}{5} = -\frac{7}{5} \):
\[ x = -2 \]

**Roots**: \( x = \frac{4}{5} \) and \( x = -2 \)

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