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Complete the square to identify the two values that solve the following quadratic equation: x2+8x=9 .(1 point) x=blank

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To complete the square for the quadratic equation \( x^2 + 8x = 9 \), follow these steps:

  1. Move the constant to the other side: \[ x^2 + 8x - 9 = 0 \]

  2. Isolate the quadratic and linear terms: \[ x^2 + 8x = 9 \]

  3. Complete the square: To complete the square, take half of the coefficient of \( x \) (which is 8), square it, and add it to both sides. Half of 8 is 4, and squaring it gives \( 4^2 = 16 \).

    \[ x^2 + 8x + 16 = 9 + 16 \]

  4. Rewrite the left side as a squared term: \[ (x + 4)^2 = 25 \]

  5. Take the square root of both sides: \[ x + 4 = \pm 5 \]

  6. Solve for \( x \):

    • For the positive case: \[ x + 4 = 5 \implies x = 5 - 4 = 1 \]
    • For the negative case: \[ x + 4 = -5 \implies x = -5 - 4 = -9 \]

Thus, the two values of \( x \) that solve the quadratic equation \( x^2 + 8x = 9 \) are: \[ x = 1 \quad \text{and} \quad x = -9 \]

So, the final answer for \( x \) is: \[ x = 1, -9 \]

Using the quadratic formula, find the solution to x2+6x+9=0 .(1 point)

To solve the quadratic equation \( x^2 + 6x + 9 = 0 \) using the quadratic formula, we first identify the coefficients \( a \), \( b \), and \( c \) from the standard form \( ax^2 + bx + c = 0 \):

  • \( a = 1 \)
  • \( b = 6 \)
  • \( c = 9 \)

The quadratic formula is given by:

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

Now, we can substitute the values of \( a \), \( b \), and \( c \) into the formula:

  1. Calculate \( b^2 - 4ac \):

\[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \]

  1. Substitute into the quadratic formula:

Since \( b^2 - 4ac = 0 \), there is exactly one solution (a double root):

\[ x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1} \]

  1. Simplify:

\[ x = \frac{-6 \pm 0}{2} = \frac{-6}{2} = -3 \]

Thus, the solution to the equation \( x^2 + 6x + 9 = 0 \) is:

\[ x = -3 \]

Solve −2x2+3x−1=−2 using the quadratic formula. Round the solution(s) to the nearest hundredths. Enter the smaller solution first.(1 point) x=blank or blank

To solve the equation \(-2x^2 + 3x - 1 = -2\) using the quadratic formula, we first rearrange it into standard form \(ax^2 + bx + c = 0\).

  1. Rearrange the equation:

\[ -2x^2 + 3x - 1 + 2 = 0 \] \[ -2x^2 + 3x + 1 = 0 \]

Now we can identify the coefficients:

  • \(a = -2\)
  • \(b = 3\)
  • \(c = 1\)
  1. Apply the quadratic formula:

The quadratic formula is:

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

  1. Calculate the discriminant:

First, compute \(b^2 - 4ac\):

\[ b^2 - 4ac = 3^2 - 4 \cdot (-2) \cdot 1 = 9 + 8 = 17 \]

  1. Use the quadratic formula:

\[ x = \frac{-3 \pm \sqrt{17}}{2 \cdot (-2)} = \frac{-3 \pm \sqrt{17}}{-4} \]

To separate the two solutions, we have:

\[ x = \frac{-3 + \sqrt{17}}{-4} \quad \text{and} \quad x = \frac{-3 - \sqrt{17}}{-4} \]

  1. Calculate the two solutions:

First solution:

\[ x_1 = \frac{-3 + \sqrt{17}}{-4} \]

Second solution:

\[ x_2 = \frac{-3 - \sqrt{17}}{-4} \]

Next, we will calculate the numerical values:

  • Calculate \( \sqrt{17} \) (approximately):

    \( \sqrt{17} \approx 4.1231 \)

  • Now substituting back:

\[ x_1 = \frac{-3 + 4.1231}{-4} = \frac{1.1231}{-4} \approx -0.2808 \]

\[ x_2 = \frac{-3 - 4.1231}{-4} = \frac{-7.1231}{-4} \approx 1.7808 \]

  1. Round the solutions to the nearest hundredths:
  • \( x_1 \approx -0.28 \)
  • \( x_2 \approx 1.78 \)

Thus, the solutions to the equation \(-2x^2 + 3x - 1 = -2\) are:

\[ x = -0.28 \quad \text{or} \quad x = 1.78 \]

Final Answer

x = -0.28 or 1.78