Question

Solve for the roots in simplest form using the quadratic formula:
4x2 +69=36x

Answer 1

To solve the equation \( 4x^2 + 69 = 36x \) using the quadratic formula, we first rearrange it into standard form \( ax^2 + bx + c = 0 \).

1. Rearranging the equation:
\[
4x^2 - 36x + 69 = 0
\]

2. Identify the coefficients \( a \), \( b \), and \( c \):
- \( a = 4 \)
- \( b = -36 \)
- \( c = 69 \)

3. Plug these coefficients into the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

4. Calculate \( b^2 - 4ac \):
\[
b^2 = (-36)^2 = 1296
\]
\[
4ac = 4 \cdot 4 \cdot 69 = 1104
\]
\[
b^2 - 4ac = 1296 - 1104 = 192
\]

5. Substitute back into the quadratic formula:
\[
x = \frac{-(-36) \pm \sqrt{192}}{2 \cdot 4}
\]
\[
x = \frac{36 \pm \sqrt{192}}{8}
\]

6. Simplify \( \sqrt{192} \):
\[
\sqrt{192} = \sqrt{64 \cdot 3} = 8\sqrt{3}
\]

7. Substitute \( \sqrt{192} \) back into the equation:
\[
x = \frac{36 \pm 8\sqrt{3}}{8}
\]
\[
x = \frac{36}{8} \pm \frac{8\sqrt{3}}{8}
\]
\[
x = \frac{9}{2} \pm \sqrt{3}
\]

Thus, the roots in simplest form are:
\[
x = \frac{9}{2} + \sqrt{3} \quad \text{and} \quad x = \frac{9}{2} - \sqrt{3}
\]