Question

The quadratic equation x2+4x−32=0 can be solved by factoring. Which of the following is the factored form?(1 point)
Responses

(x−4)(x−8)=0
left parenthesis x minus 4 right parenthesis left parenthesis x minus 8 right parenthesis equals 0

(x+4)(x+8)=0
left parenthesis x plus 4 right parenthesis left parenthesis x plus 8 right parenthesis equals 0

(x+8)(x−4)=0
left parenthesis x plus 8 right parenthesis left parenthesis x minus 4 right parenthesis equals 0

(x+4)(x−8)=0

Answer 1

Which of the following equations has exactly the same solutions as (x−7)(x+9)=0 ?(1 point)
Responses

x2=−2x+63
x squared equals negative 2 x plus 63

x2−2x−63=0
x squared minus 2 x minus 63 equals 0

x2+2x=−63
x squared plus 2 x equals negative 63

x2=2x−63

Answer 2

What should be the value of b in x2+bx−55=0 so that its solutions are −11 and 5? (1 point)
Responses

−16
negative 16

−6
negative 6

6
6

16

Answer 3

Solve the following quadratic equation using the perfect square trinomial pattern: x2−16x+64=0.(1 point)
Responses

x=−64
x equals negative 64

x=64
x equals 64

x=8
x equals 8

x=−8

Answer 4

To factor the quadratic equation \(x^2 + 4x - 32 = 0\), we need to find two numbers that multiply to \(-32\) (the constant term) and add to \(4\) (the coefficient of the linear term).

The two numbers that satisfy these conditions are \(8\) and \(-4\) because:

- \(8 \times -4 = -32\)
- \(8 + (-4) = 4\)

Thus, we can factor the quadratic as:

\[
(x + 8)(x - 4) = 0
\]

So, the correct factored form is:

\[
\boxed{(x + 8)(x - 4) = 0}
\]

Answer 5

The equation \((x - 7)(x + 9) = 0\) has the solutions \(x = 7\) and \(x = -9\). We can find the equivalent equation that also has these solutions by expanding the factored form:

1. **Expand \((x - 7)(x + 9)\)**:
\[
x^2 + 9x - 7x - 63 = x^2 + 2x - 63
\]
So the equation can also be written as:
\[
x^2 + 2x - 63 = 0
\]

Now, let's look at the provided options:

1. \(x^2 = -2x + 63\)
2. \(x^2 - 2x - 63 = 0\)
3. \(x^2 + 2x = -63\)
4. \(x^2 = 2x - 63\)

To determine which one has the same solutions, we will rewrite some of these options:

- **Option 1**: Rearranging gives \(x^2 + 2x - 63 = 0\).
- **Option 2**: This is already in standard form.
- **Option 3**: Rearranging gives \(x^2 + 2x + 63 = 0\) (not the same).
- **Option 4**: Rearranging gives \(x^2 - 2x + 63 = 0\) (not the same).

So, the equations \(x^2 - 2x - 63 = 0\) and \(x^2 + 2x - 63 = 0\) need to be evaluated further. Let's focus back on the correct factors:

The equation that correctly simplifies down to the same roots as \((x - 7)(x + 9) = 0\) is **Option 2**:

\[
\boxed{x^2 - 2x - 63 = 0}
\]

This equation indeed has the same solutions.

Answer 6

To find the value of \( b \) in the equation \( x^2 + bx - 55 = 0 \) such that the solutions (roots) are \( x = -11 \) and \( x = 5 \), we can use Vieta's formulas. According to Vieta's, for a quadratic equation of the form \( x^2 + bx + c = 0 \):

1. The sum of the roots \( r_1 \) and \( r_2 \) is given by \( r_1 + r_2 = -b \).
2. The product of the roots is given by \( r_1 \times r_2 = c \).

Given the roots are \( -11 \) and \( 5 \):

1. **Sum of the roots**:
\[
-11 + 5 = -6
\]
Therefore, according to Vieta's, we have:
\[
-b = -6 \implies b = 6
\]

2. **Product of the roots** (for confirmation):
\[
-11 \times 5 = -55
\]
This matches the constant term in the equation.

Thus, the value of \( b \) should be \( 6 \).

The correct response is:

\[
\boxed{6}
\]

Answer 7

The given quadratic equation is

\[
x^2 - 16x + 64 = 0.
\]

To solve this equation using the perfect square trinomial pattern, we first recognize that it can be expressed as a square of a binomial.

The expression \( x^2 - 16x + 64 \) can be rewritten as:

\[
(x - 8)^2 = 0.
\]

Next, we solve for \( x \) by taking the square root of both sides:

\[
x - 8 = 0.
\]

This simplifies to:

\[
x = 8.
\]

Thus, the solution to the equation is:

\[
\boxed{x = 8}.
\]