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Complete the square to identify the two values that solve the following quadratic equation: x^2+8x=9 .(1 point) x=()

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

  1. Rearrange the equation to set it to zero: \[ x^2 + 8x - 9 = 0 \]

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

  3. 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 \( 4^2 = 16 \): \[ x^2 + 8x + 16 = 9 + 16 \] This simplifies to: \[ x^2 + 8x + 16 = 25 \]

  4. Now, the left-hand side can be factored as a perfect square: \[ (x + 4)^2 = 25 \]

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

  6. Solve for \( x \) by isolating it: \[ x + 4 = 5 \quad \Rightarrow \quad x = 1 \] \[ x + 4 = -5 \quad \Rightarrow \quad x = -9 \]

Thus, the solutions to the equation \( x^2 + 8x = 9 \) are: \[ \boxed{1} \quad \text{and} \quad \boxed{-9} \]

Complete the square to solve the following quadratic equation: x^2+4x−6=2 . Is the following correct: x=−2+23–√, −2−23–√ ? Enter 1 for Yes. Enter 2 for No. (1 point)

Let's solve the equation \( x^2 + 4x - 6 = 2 \) by completing the square step by step.

  1. First, rearrange the equation to set it to zero: \[ x^2 + 4x - 6 - 2 = 0 \quad \Rightarrow \quad x^2 + 4x - 8 = 0 \]

  2. Move the constant term to the other side: \[ x^2 + 4x = 8 \]

  3. To complete the square, take half of the coefficient of \( x \) (which is 4), square it, and add it to both sides. Half of 4 is 2, and squaring it gives \( 2^2 = 4 \): \[ x^2 + 4x + 4 = 8 + 4 \] This simplifies to: \[ (x + 2)^2 = 12 \]

  4. Take the square root of both sides: \[ x + 2 = \pm \sqrt{12} \]

  5. Isolate \( x \): \[ x = -2 \pm \sqrt{12} \] Since \( \sqrt{12} = 2\sqrt{3} \), we have: \[ x = -2 \pm 2\sqrt{3} \]

Now, we compare this with the provided answer \( x = -2 + 23 - \sqrt{} \) and \( x = -2 - 23 - \sqrt{} \).

The correct values are: \[ x = -2 + 2\sqrt{3} \quad \text{and} \quad x = -2 - 2\sqrt{3} \]

Since the values given in the question do not match the correct values derived from completing the square, the answer is:

Enter 2 for No.

Using the quadratic formula, find the solution to x^2+6x+9=0 .(1 point)

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

In this case:

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

The quadratic formula is given by:

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

Now, let's substitute the coefficients into the formula:

  1. Calculate the discriminant: \[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \]

  2. Substitute \( a \), \( b \), and the discriminant into the quadratic formula: \[ x = \frac{-6 \pm \sqrt{0}}{2 \cdot 1} \]

  3. Since the square root of 0 is 0, the equation simplifies to: \[ x = \frac{-6 \pm 0}{2} = \frac{-6}{2} = -3 \]

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

\[ x = -3 \]

Since this is a double root (the discriminant is 0), the solution can be expressed as: \[ \boxed{-3} \]

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

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

  1. Rearranging the equation: \[ -2x^2 + 3x - 1 + 2 = 0 \quad \Rightarrow \quad -2x^2 + 3x + 1 = 0 \]

  2. We can multiply through by -1 to simplify: \[ 2x^2 - 3x - 1 = 0 \]

Now, we identify the coefficients \(a\), \(b\), and \(c\):

  • \( a = 2 \)
  • \( b = -3 \)
  • \( c = -1 \)

  1. Now, use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

  2. Calculate the discriminant \(b^2 - 4ac\): \[ b^2 - 4ac = (-3)^2 - 4(2)(-1) = 9 + 8 = 17 \]

  3. Substitute into the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{17}}{2 \cdot 2} \] \[ x = \frac{3 \pm \sqrt{17}}{4} \]

  4. Now we compute the two solutions:

    • For the positive case: \[ x_1 = \frac{3 + \sqrt{17}}{4} \]
    • For the negative case: \[ x_2 = \frac{3 - \sqrt{17}}{4} \]

  5. Calculate the approximate values:

    • \(\sqrt{17} \approx 4.123\)
    • Calculate \(x_1\): \[ x_1 \approx \frac{3 + 4.123}{4} = \frac{7.123}{4} \approx 1.78075 \quad \rightarrow \quad 1.78 \text{ (rounded to nearest hundredths)} \]
    • Calculate \(x_2\): \[ x_2 \approx \frac{3 - 4.123}{4} = \frac{-1.123}{4} \approx -0.28075 \quad \rightarrow \quad -0.28 \text{ (rounded to nearest hundredths)} \]

Thus, the solutions, with the smaller solution first, are:

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

So the final answer is:

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

How many real solutions does the following quadratic equation have? 25x^2+60x+36=0 (1 point) ()solution(s)

To determine how many real solutions the quadratic equation \(25x^2 + 60x + 36 = 0\) has, we need to calculate the discriminant, which is given by the formula:

\[ D = b^2 - 4ac \]

Here, the coefficients are:

  • \(a = 25\)
  • \(b = 60\)
  • \(c = 36\)

Now, let's calculate the discriminant:

  1. Calculate \(b^2\): \[ b^2 = 60^2 = 3600 \]

  2. Calculate \(4ac\): \[ 4ac = 4 \cdot 25 \cdot 36 = 3600 \]

  3. Now compute the discriminant: \[ D = b^2 - 4ac = 3600 - 3600 = 0 \]

The discriminant \(D = 0\) indicates that there is exactly one real solution (a double root) for the quadratic equation.

Therefore, the answer is:

\[ \text{1 solution} \]