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The total number of seats in an auditorium is modeled by `f\left(x\right)=2x^{2}-6x,`where `x` represents the number of rows of seats. How many rows are there n the auditorium if it has a total of 416 seats?

11 answers

To find the number of rows in the auditorium given that the total number of seats is 416, we need to set the function equal to 416 and solve for \( x \):

\[ f(x) = 2x^2 - 6x = 416 \]

First, we'll rewrite the equation:

\[ 2x^2 - 6x - 416 = 0 \]

Next, we can simplify this equation by dividing all terms by 2:

\[ x^2 - 3x - 208 = 0 \]

Now, we can solve this quadratic equation using the quadratic formula:

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

In our equation, \( a = 1 \), \( b = -3 \), and \( c = -208 \). We first calculate the discriminant:

\[ b^2 - 4ac = (-3)^2 - 4 \cdot 1 \cdot (-208) = 9 + 832 = 841 \]

Now we can find the roots using the quadratic formula:

\[ x = \frac{-(-3) \pm \sqrt{841}}{2 \cdot 1} = \frac{3 \pm 29}{2} \]

Calculating the two possible values for \( x \):

  1. \( x = \frac{3 + 29}{2} = \frac{32}{2} = 16 \)
  2. \( x = \frac{3 - 29}{2} = \frac{-26}{2} = -13 \)

Since \( x \) represents the number of rows of seats, it must be a non-negative integer. Therefore, we discard \( x = -13 \).

Thus, the number of rows in the auditorium is:

\[ \boxed{16} \]

What are the solutions to `\left(x+7\right)^{2}=81?`

To solve the equation \( (x + 7)^2 = 81 \), we start by taking the square root of both sides. Remembering to consider both the positive and negative square roots, we get:

\[ x + 7 = \pm 9 \]

This gives us two separate equations to solve:

  1. \( x + 7 = 9 \)
  2. \( x + 7 = -9 \)

For the first equation: \[ x + 7 = 9 \] Subtracting 7 from both sides gives: \[ x = 9 - 7 = 2 \]

For the second equation: \[ x + 7 = -9 \] Subtracting 7 from both sides gives: \[ x = -9 - 7 = -16 \]

Thus, the solutions to the equation \( (x + 7)^2 = 81 \) are:

\[ x = 2 \quad \text{and} \quad x = -16 \]

So the final solutions are:

\[ \boxed{2 \text{ and } -16} \]

What are the solutions to `2\left(x-7\right)^{2}=32?`

To solve the equation \( 2\left(x - 7\right)^{2} = 32 \), we'll start by isolating the squared term.

First, divide both sides by 2:

\[ \left(x - 7\right)^{2} = \frac{32}{2} \]

This simplifies to:

\[ \left(x - 7\right)^{2} = 16 \]

Next, we take the square root of both sides, remembering to consider both the positive and negative roots:

\[ x - 7 = \pm 4 \]

This gives us two separate equations to solve:

  1. \( x - 7 = 4 \)
  2. \( x - 7 = -4 \)

For the first equation: \[ x - 7 = 4 \] Adding 7 to both sides gives: \[ x = 4 + 7 = 11 \]

For the second equation: \[ x - 7 = -4 \] Adding 7 to both sides gives: \[ x = -4 + 7 = 3 \]

Thus, the solutions to the equation \( 2\left(x - 7\right)^{2} = 32 \) are:

\[ x = 11 \quad \text{and} \quad x = 3 \]

So the final solutions are:

\[ \boxed{11 \text{ and } 3} \]

Suppose you are a solving a quadratic equation using the quadratic formula and this is your work so far:

It seems like you may have been about to ask a question regarding solving a quadratic equation or you may have wanted to provide details of your work so far. If you're working with a specific quadratic equation and the quadratic formula, please provide the equation you've been dealing with, and I'll be happy to help you continue from where you left off or answer your questions!

The quadratic formula is given by:

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

where \( ax^2 + bx + c = 0 \) is the standard form of the quadratic equation. If you can share the values of \( a \), \( b \), and \( c \), or specify what you've done up to this point, I can assist you further!

Find the values of a, b, and c for the quadratic equation below (remember `ax^{2}+bx+c=0`)

`x^{2}-7x=4`

To write the given equation \( x^2 - 7x = 4 \) in the standard form \( ax^2 + bx + c = 0 \), we first need to move all terms to one side of the equation.

Starting with:

\[ x^2 - 7x - 4 = 0 \]

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

  • \( a = 1 \) (coefficient of \( x^2 \))
  • \( b = -7 \) (coefficient of \( x \))
  • \( c = -4 \) (constant term)

Thus, we have:

\[ \boxed{a = 1, ; b = -7, ; c = -4} \]

What are the solutions to `x^{2}+32x+248\ =\ 0`?

To find the solutions to the quadratic equation \( x^2 + 32x + 248 = 0 \), we can use the quadratic formula:

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

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

  • \( a = 1 \)
  • \( b = 32 \)
  • \( c = 248 \)

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

\[ b^2 - 4ac = 32^2 - 4 \cdot 1 \cdot 248 \] \[ b^2 - 4ac = 1024 - 992 = 32 \]

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

\[ x = \frac{-32 \pm \sqrt{32}}{2 \cdot 1} \] \[ x = \frac{-32 \pm 4\sqrt{2}}{2} \] \[ x = \frac{-32}{2} \pm \frac{4\sqrt{2}}{2} \] \[ x = -16 \pm 2\sqrt{2} \]

Therefore, the two solutions are:

\[ x = -16 + 2\sqrt{2} \quad \text{and} \quad x = -16 - 2\sqrt{2} \]

So, the final solutions are:

\[ \boxed{-16 + 2\sqrt{2}} \quad \text{and} \quad \boxed{-16 - 2\sqrt{2}} \]