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

10±25√5i
the fraction with numerator 10 plus or minus 2 square root of 5 and denominator 5 i

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

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

10±10√5i

Answer 1

To solve the quadratic equation \( 10x^2 + 5x = 5x - 8 \), we first simplify it. Subtract \( 5x \) from both sides:

\[
10x^2 + 5x - 5x = -8
\]

This simplifies to:

\[
10x^2 = -8
\]

Next, we can rearrange this equation:

\[
10x^2 + 8 = 0
\]

Now, we can rewrite this in standard form, \( ax^2 + bx + c = 0 \):

\[
10x^2 + 0x + 8 = 0
\]

Here, \( a = 10 \), \( b = 0 \), and \( c = 8 \).

We will use the quadratic formula:

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

Substituting our values for \( a \), \( b \), and \( c \):

\[
x = \frac{-0 \pm \sqrt{0^2 - 4(10)(8)}}{2(10)}
\]

Calculating the discriminant:

\[
0^2 - 4(10)(8) = 0 - 320 = -320
\]

Now we substitute back into the formula:

\[
x = \frac{0 \pm \sqrt{-320}}{20}
\]

Next, we simplify \( \sqrt{-320} \):

\[
\sqrt{-320} = \sqrt{320} \cdot i = \sqrt{64 \cdot 5} \cdot i = 8\sqrt{5} \cdot i
\]

Now substitute this into the formula:

\[
x = \frac{0 \pm 8\sqrt{5} i}{20}
\]

This simplifies to:

\[
x = \frac{8\sqrt{5} i}{20} = \frac{2\sqrt{5} i}{5}
\]

Thus, our final solutions are:

\[
x = 0 \pm \frac{2\sqrt{5}}{5} i
\]

The correct response from your options would be:

\( 0 \pm \frac{2\sqrt{5}}{5} i \)