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Using the quadratic formula, find the solution to x2+6x+9=0

.(1 point)

11 answers

To find the solutions of the quadratic equation \( x^2 + 6x + 9 = 0 \) using the quadratic formula, we start by identifying the coefficients \( a \), \( b \), and \( c \):

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

The quadratic formula is given by:

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

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

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

Since the discriminant is \( 0 \), we will have one real root. Now substituting into the quadratic formula:

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

This simplifies to:

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

Thus, we find:

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

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

\[ \boxed{-3} \]

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

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

Starting with the equation:

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

This simplifies to:

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

Now, we identify the coefficients:

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

Next, we apply the quadratic formula:

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

We first calculate the discriminant \(b^2 - 4ac\):

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

Now substituting back into the quadratic formula:

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

This becomes:

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

Now we separate the two possible solutions:

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

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

Calculating the numerical values for these solutions:

  1. Calculate \(\sqrt{17}\):

    \(\sqrt{17} \approx 4.123\)

So:

\[ x_1 = \frac{-3 + 4.123}{-4} = \frac{1.123}{-4} \approx -0.28075 \text{ (rounding gives } -0.28 \text{ )} \]

\[ x_2 = \frac{-3 - 4.123}{-4} = \frac{-7.123}{-4} \approx 1.78075 \text{ (rounding gives } 1.78 \text{ )} \]

Thus the solutions, rounded to the nearest hundredths, are:

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

In conclusion, the solutions are:

\[ \boxed{-0.28} \quad \text{and} \quad \boxed{1.78} \]

How many real solutions does the following quadratic equation have?

25x2+60x+36=0

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

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

where \(a\), \(b\), and \(c\) are the coefficients from the standard form \(ax^2 + bx + c = 0\).

For the given equation:

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

Now, substituting these values into the discriminant formula:

\[ D = 60^2 - 4 \cdot 25 \cdot 36 \]

Calculating \(60^2\):

\[ 60^2 = 3600 \]

Now calculating \(4 \cdot 25 \cdot 36\):

\[ 4 \cdot 25 = 100 \] \[ 100 \cdot 36 = 3600 \]

Now putting it all together:

\[ D = 3600 - 3600 = 0 \]

The value of the discriminant \(D\) is \(0\).

Conclusion:

When the discriminant is:

  • \(D > 0\): There are two distinct real solutions.
  • \(D = 0\): There is exactly one real solution (a double root).
  • \(D < 0\): There are no real solutions.

Since \(D = 0\) for this equation, it has exactly one real solution.

Question
Complete the square to solve the following quadratic equation: x2+2x−8=0
.(1 point)
Responses

x=−10
, x=8
x equals negative 10 , x equals 8

x=10
, x=−8
x equals 10 , x equals negative 8

x=2
, x=−4
x equals 2 , x equals negative 4

x=−2
, x=4

To solve the quadratic equation \( x^2 + 2x - 8 = 0 \) by completing the square, we will follow these steps:

  1. Move the constant term to the other side of the equation:

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

  2. Complete the square:

    • Take half of the coefficient of \(x\), which is \(2\). Half of \(2\) is \(1\), and squaring it gives \(1^2 = 1\).
    • Add and subtract this square (\(1\)) inside the equation.

    \[ x^2 + 2x + 1 - 1 = 8 \]

    This simplifies to:

    \[ (x + 1)^2 - 1 = 8 \]

    Now add \(1\) to both sides:

    \[ (x + 1)^2 = 9 \]

  3. Take the square root of both sides:

    \[ x + 1 = \pm 3 \]

  4. Solve for \(x\):

    • First solution:

    \[ x + 1 = 3 \implies x = 3 - 1 = 2 \]

    • Second solution:

    \[ x + 1 = -3 \implies x = -3 - 1 = -4 \]

Thus, the solutions are:

\[ x = 2 \quad \text{and} \quad x = -4 \]

Rounding off to the options given, the correct response is:

\[ \boxed{x = 2, ; x = -4} \]

Complete the square to solve the following quadratic equation: x2−22x+1=15 .(1 point)
Responses

x=11+14−−√, 11−14−−√
x equals 11 plus Start Root 14 End Root comma 11 minus Start Root 14 End Root

x=−11+14−−√, −11−14−−√
x equals negative 11 plus Start Root 14 End Root comma negative 11 minus Start Root 14 End Root

x=−11+315−−√, −11−315−−√
x equals negative 11 plus 3 Start Root 15 End Root comma negative 11 minus 3 Start Root 15 End Root

x=11+315−−√, 11−315−−√

To solve the quadratic equation \(x^2 - 22x + 1 = 15\) by completing the square, we will follow these steps:

  1. Rearrange the equation: Move \(15\) to the left side:

    \[ x^2 - 22x + 1 - 15 = 0 \]

    Simplifying gives:

    \[ x^2 - 22x - 14 = 0 \]

  2. Move the constant to the other side: \[ x^2 - 22x = 14 \]

  3. Complete the square:

    • Take half of the coefficient of \(x\) (which is \(-22\)), so half of \(-22\) is \(-11\), and squaring it gives \((-11)^2 = 121\).
    • Add and subtract this square:

    \[ x^2 - 22x + 121 - 121 = 14 \]

    This can be rewritten as:

    \[ (x - 11)^2 - 121 = 14 \]

    Now add \(121\) to both sides:

    \[ (x - 11)^2 = 14 + 121 = 135 \]

  4. Take the square root of both sides:

    \[ x - 11 = \pm \sqrt{135} \]

  5. Solve for \(x\):

    \[ x = 11 \pm \sqrt{135} \]

This gives us two solutions:

\[ x = 11 + \sqrt{135} \quad \text{and} \quad x = 11 - \sqrt{135} \]

Final Answer:

Now, reducing \(\sqrt{135}\):

\[ \sqrt{135} = \sqrt{9 \cdot 15} = 3\sqrt{15} \]

So the solutions can also be written as:

\[ x = 11 + 3\sqrt{15} \quad \text{and} \quad x = 11 - 3\sqrt{15} \]

The correct response is:

\[ \boxed{x = 11 + 3\sqrt{15}, ; 11 - 3\sqrt{15}} \]

Using the quadratic formula, find the solution to x2−6x−16=0 .(1 point)
Responses

x=8, x=−2
x equals 8 , x equals negative 2

x=8
x equals 8

x=−2
x equals negative 2

x=8, x=−8

To solve the quadratic equation \(x^2 - 6x - 16 = 0\) using the quadratic formula, we start by identifying the coefficients \(a\), \(b\), and \(c\):

  • \(a = 1\)
  • \(b = -6\)
  • \(c = -16\)

The quadratic formula is given by:

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

Step 1: Calculate the Discriminant

We first calculate the discriminant \(D\):

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

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

\[ D = (-6)^2 - 4 \cdot 1 \cdot (-16) \] \[ D = 36 + 64 \] \[ D = 100 \]

Step 2: Apply the Quadratic Formula

Now substitute the values of \(b\), \(D\), and \(a\) into the quadratic formula:

\[ x = \frac{-(-6) \pm \sqrt{100}}{2 \cdot 1} \] \[ x = \frac{6 \pm 10}{2} \]

Step 3: Solve for \(x\)

This gives us two potential solutions:

  1. For the positive case: \[ x_1 = \frac{6 + 10}{2} = \frac{16}{2} = 8 \]

  2. For the negative case: \[ x_2 = \frac{6 - 10}{2} = \frac{-4}{2} = -2 \]

Final Solutions

Thus, the two solutions to the equation \(x^2 - 6x - 16 = 0\) are:

\[ x = 8 \quad \text{and} \quad x = -2 \]

Correct Response

The correct response is:

\[ \boxed{x = 8, ; x = -2} \]