Question

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

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 it into standard form by moving all terms to one side:

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

Next, we identify the coefficients \(a\), \(b\), and \(c\):

- \(a = 8\)
- \(b = -12\)
- \(c = 5\)

We will use the quadratic formula:

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

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

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

Since the discriminant is negative, we will have complex solutions.

Now, substituting the values into the quadratic formula:

\[
x = \frac{-(-12) \pm \sqrt{-16}}{2 \cdot 8}
\]
\[
x = \frac{12 \pm \sqrt{-16}}{16}
\]
\[
x = \frac{12 \pm 4i}{16}
\]
\[
x = \frac{12}{16} \pm \frac{4i}{16}
\]
\[
x = \frac{3}{4} \pm \frac{1}{4}i
\]

Thus, the solutions are:

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

Now we compare this with the given options:

A: \(\frac{3}{4} \pm \frac{1}{4}i\) (which is correct)
B: \(12 \pm 4i\)
C: \(12 \pm \frac{1}{4}i\)
D: \(-\frac{3}{4} + \frac{1}{4}i\)

The correct answer is:

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