Use the quadratic formula to solve the quadratic equation 10x^2+5x=5x-8. Express its solutions in the form a± bi. (1 point)

1 answer

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

Subtract \(5x\) from both sides:

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

Next, rearranging gives:

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

Now, let's bring -8 to the left side:

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

We identify the coefficients:

  • \(a = 10\)
  • \(b = 0\)
  • \(c = 8\)

Now, we can use the quadratic formula, which is:

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

Plugging in the values:

  1. Calculate \(b^2 - 4ac\): \[ b^2 - 4ac = 0^2 - 4(10)(8) = 0 - 320 = -320 \]

  2. Substitute into the quadratic formula: \[ x = \frac{-0 \pm \sqrt{-320}}{2(10)} = \frac{\pm \sqrt{-320}}{20} \]

  3. Simplifying \(\sqrt{-320}\): \[ \sqrt{-320} = \sqrt{320} \cdot i = \sqrt{64 \cdot 5} \cdot i = 8\sqrt{5} \cdot i \]

  4. Thus: \[ x = \frac{\pm 8\sqrt{5} \cdot i}{20} = \frac{8\sqrt{5} \cdot i}{20} = \frac{2\sqrt{5} \cdot i}{5} \]

So, the solutions are:

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

In standard form:

\[ x = 0 + \frac{2\sqrt{5}}{5} i \quad \text{and} \quad x = 0 - \frac{2\sqrt{5}}{5} i \]

Thus, the final solutions are:

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