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Complete the square to identify the two values that solve the following quadratic equation: x2+8x=9

.(1 point)
x=

7 answers

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

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

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

  3. 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 \( 4^2 = 16 \).

    Add 16 to both sides: \[ x^2 + 8x + 16 = 9 + 16 \] \[ x^2 + 8x + 16 = 25 \]

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

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

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 start with the general form of a quadratic equation:

\[ ax^2 + bx + c = 0 \]

For the given equation:

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

The quadratic formula is given by:

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

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

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

  2. Plug values into the quadratic formula: \[ x = \frac{{-6 \pm \sqrt{0}}}{{2(1)}} \] \[ x = \frac{{-6 \pm 0}}{2} = \frac{{-6}}{2} = -3 \]

Since the discriminant is 0, there is only one real solution.

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=
or

To solve the equation \(-2x^2 + 3x - 1 = -2\) using the quadratic formula, first rearrange the equation to set it to zero:

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

This simplifies to:

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

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

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

The quadratic formula is given by:

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

Substituting in the values of \(a\), \(b\), and \(c\):

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

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

  3. Evaluate the two potential solutions:

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

  4. Calculate the numeric values:

    First, calculate \(\sqrt{17} \approx 4.123\).

    Now plug this value into the formulas:

    \[ x_1 = \frac{{-3 + 4.123}}{{-4}} \approx \frac{{1.123}}{{-4}} \approx -0.28075 \quad (\text{approximately } -0.28) \] \[ x_2 = \frac{{-3 - 4.123}}{{-4}} \approx \frac{{-7.123}}{{-4}} \approx 1.78075 \quad (\text{approximately } 1.78) \]

Therefore, rounding both solutions to the nearest hundredths, we find:

  • Smaller solution: \( x \approx -0.28 \)
  • Larger solution: \( x \approx 1.78 \)

Thus, the two solutions are: \[ x \approx -0.28 \quad \text{or} \quad x \approx 1.78 \]

How many real solutions does the following quadratic equation have?

25x2+60x+36=0

(1 point)
solution(s)

To determine how many real solutions the quadratic equation \( 25x^2 + 60x + 36 = 0 \) has, we can calculate the discriminant \( D \).

The discriminant is given by the formula:

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

For the given equation, identify the coefficients:

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

Substituting these values into the discriminant formula:

\[ D = 60^2 - 4(25)(36) \] \[ D = 3600 - 3600 \] \[ D = 0 \]

The discriminant \( D = 0 \) indicates that the quadratic equation has exactly one real solution (also called a repeated root).

Thus, the answer is:

  • One real solution.