Question

Use the quadratic formula to solve the quadratic equation 10x2+5x=5x−8
. Express its solutions in the form a+bi
(1 point)
Responses

0±5√10i
0 plus or minus the fraction with numerator square root of 5 and denominator 10 i

2±5i
2 plus or minus 5 i

0±25√5i
0 plus or minus the fraction with numerator 2 square root of 5 and denominator 5 i

5±25i

Answer 1

To solve the quadratic equation \(10x^2 + 5x = 5x - 8\), we first rearrange it into the standard form \(ax^2 + bx + c = 0\).

1. Subtract \(5x\) from both sides:
\[
10x^2 + 5x - 5x = -8
\]
Simplifying this gives:
\[
10x^2 = -8
\]

2. Add \(8\) to both sides to get:
\[
10x^2 + 8 = 0
\]

3. Rearranging to:
\[
10x^2 + 0x + 8 = 0
\]

Now we can identify the coefficients \(a\), \(b\), and \(c\):
- \(a = 10\)
- \(b = 0\)
- \(c = 8\)

4. We apply the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[
x = \frac{-0 \pm \sqrt{0^2 - 4 \cdot 10 \cdot 8}}{2 \cdot 10}
\]
\[
x = \frac{\pm \sqrt{0 - 320}}{20}
\]
\[
x = \frac{\pm \sqrt{-320}}{20}
\]

5. Since we have a negative number under the square root, we can simplify:
\[
\sqrt{-320} = \sqrt{320}i = \sqrt{64 \cdot 5}i = 8\sqrt{5}i
\]
Thus:
\[
x = \frac{\pm 8\sqrt{5}i}{20}
\]
\[
x = \frac{\pm 2\sqrt{5}i}{5}
\]

So the final solutions are:
\[
x = 0 \pm \frac{2\sqrt{5}}{5}i
\]

To convert to standard format \(a + bi\), we express it as:
\[
x = 0 + \frac{2\sqrt{5}}{5}i \quad \text{and} \quad x = 0 - \frac{2\sqrt{5}}{5}i
\]

The closest answer option aligning with this result is:
- \(0 \pm \frac{2\sqrt{5}}{5}i\), which corresponds to the option "0 plus or minus the fraction with numerator 2 square root of 5 and denominator 5 i".