solve the equation x, squared, plus, 2, x, minus, 17, equals, 0x

2
+2x−17=0 to the nearest tenth.

1 answer

To solve the equation \( x^2 + 2x - 17 = 0 \), we can use the quadratic formula:

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

In this equation, \( a = 1 \), \( b = 2 \), and \( c = -17 \).

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

\[ b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot (-17) = 4 + 68 = 72 \]

Now, we can substitute the values into the quadratic formula:

\[ x = \frac{{-2 \pm \sqrt{72}}}{2 \cdot 1} \]

Calculating \( \sqrt{72} \):

\[ \sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2} \approx 6 \cdot 1.414 \approx 8.485 \]

Now substitute this back into the formula:

\[ x = \frac{{-2 \pm 8.485}}{2} \]

This gives us two possible solutions:

  1. \( x = \frac{{-2 + 8.485}}{2} = \frac{6.485}{2} \approx 3.2425 \)
  2. \( x = \frac{{-2 - 8.485}}{2} = \frac{-10.485}{2} \approx -5.2425 \)

Rounding these to the nearest tenth:

  1. \( x \approx 3.2 \)
  2. \( x \approx -5.2 \)

Thus, the solutions to the equation \( x^2 + 2x - 17 = 0 \) are approximately:

\[ x \approx 3.2 \quad \text{and} \quad x \approx -5.2 \]