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

1 answer

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