Question

Use the quadratic formula to solve the quadratic equation 8x^2+5=12x
. Express its solutions in the form a+-bi
.

A
34±14i\frac{3}{4}\pm\frac{1}{4}i
4
3
​
±
4
1
​
i

B
12±4i12\pm4i12±4i

C
12±14i12\pm\frac{1}{4}i12±
4
1
​
i

D
−34+14i-\frac{3}{4}+\frac{1}{4}i−
4
3
​
+
4
1
​
i

Answer 1

To solve the quadratic equation \( 8x^2 + 5 = 12x \), we first rearrange the equation to standard form:

\[
8x^2 - 12x + 5 = 0
\]

Now, we can use the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In this case, the coefficients are:
- \( a = 8 \)
- \( b = -12 \)
- \( c = 5 \)

Now, we calculate \( b^2 - 4ac \):

\[
b^2 = (-12)^2 = 144
\]
\[
4ac = 4 \cdot 8 \cdot 5 = 160
\]
\[
b^2 - 4ac = 144 - 160 = -16
\]

Since \( b^2 - 4ac = -16 \), we have a negative discriminant, which means the solutions will be complex.

Next, substitute the values into the quadratic formula:

\[
x = \frac{-(-12) \pm \sqrt{-16}}{2 \cdot 8}
\]
\[
x = \frac{12 \pm \sqrt{-16}}{16}
\]
\[
x = \frac{12 \pm 4i}{16}
\]

Now simplify the expression:

\[
x = \frac{12}{16} \pm \frac{4i}{16}
\]
\[
x = \frac{3}{4} \pm \frac{1}{4}i
\]

Thus, the solutions to the equation \( 8x^2 - 12x + 5 = 0 \) are:

\[
x = \frac{3}{4} \pm \frac{1}{4}i
\]

Now, looking for the solution in the provided options, we see that the correct expression corresponds to:

\[
\frac{3}{4} \pm \frac{1}{4}i
\]

This matches with option **A**:

\[
\frac{3}{4} \pm \frac{1}{4}i
\]

So the final answer is:

**A: \(\frac{3}{4} \pm \frac{1}{4}i\)**