Question

9 points) x2+6x+9=0 x= 16x2−49=0 Keep answer in fraction form. List smaller solution first. x= and 3x2+18x+24=0 List smaller solution first. x= and 2x2+x=5 Answer to the hundredths. List smaller solution first. x= and Show how to find the zero(s): y=2x2+7x List smaller solution first.

Answer 1

Let's solve each of the equations step by step.

### 1. Solve \( x^2 + 6x + 9 = 0 \)

This is a perfect square trinomial:

\[
(x + 3)^2 = 0
\]

Taking the square root of both sides, we find:

\[
x + 3 = 0 \implies x = -3
\]

### 2. Solve \( 16x^2 - 49 = 0 \)

This can be factored as a difference of squares:

\[
(4x - 7)(4x + 7) = 0
\]

Setting each factor to zero gives us:

\[
4x - 7 = 0 \implies 4x = 7 \implies x = \frac{7}{4}
\]
\[
4x + 7 = 0 \implies 4x = -7 \implies x = -\frac{7}{4}
\]

Listing smaller solution first:

\[
x = -\frac{7}{4}, \frac{7}{4}
\]

### 3. Solve \( 3x^2 + 18x + 24 = 0 \)

First, simplify the equation by dividing all terms by 3:

\[
x^2 + 6x + 8 = 0
\]

Now, factor the quadratic:

\[
(x + 2)(x + 4) = 0
\]

Setting each factor to zero gives us:

\[
x + 2 = 0 \implies x = -2
\]
\[
x + 4 = 0 \implies x = -4
\]

Listing smaller solution first (remembering we should reverse to smaller first):

\[
x = -4, -2
\]

### 4. Solve \( 2x^2 + x = 5 \)

Rearranging the equation:

\[
2x^2 + x - 5 = 0
\]

Now apply the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2, b = 1, c = -5 \):

Calculating the discriminant:

\[
b^2 - 4ac = 1^2 - 4 \cdot 2 \cdot (-5) = 1 + 40 = 41
\]

Now substitute into the quadratic formula:

\[
x = \frac{-1 \pm \sqrt{41}}{4}
\]

Calculating the two solutions:

1. \( x_1 = \frac{-1 + \sqrt{41}}{4} \)
2. \( x_2 = \frac{-1 - \sqrt{41}}{4} \)

Calculating \( x_2 \) in decimal to the hundredths:

\[
\sqrt{41} \approx 6.403 \implies x_2 \approx \frac{-1 - 6.403}{4} \approx \frac{-7.403}{4} \approx -1.85075 \approx -1.85
\]

So, listing smaller solution first, we have:

\[
x = \frac{-1 - \sqrt{41}}{4} \text{ (around -1.85)}, \frac{-1 + \sqrt{41}}{4}
\]

### 5. Finding zeros of \( y = 2x^2 + 7x \)

Set \( y = 0 \):

\[
2x^2 + 7x = 0
\]

Factoring out \( x \) gives us:

\[
x(2x + 7) = 0
\]

Setting each factor to zero gives us:

1. \( x = 0 \)
2. \( 2x + 7 = 0 \implies 2x = -7 \implies x = -\frac{7}{2} \)

Listing smaller solution first:

\[
x = -\frac{7}{2}, 0
\]

### Summary of Solutions

1. \( x = -3 \)
2. \( x = -\frac{7}{4}, \frac{7}{4} \)
3. \( x = -4, -2 \)
4. \( x = \frac{-1 - \sqrt{41}}{4}, \frac{-1 + \sqrt{41}}{4} \approx -1.85 \)
5. \( x = -\frac{7}{2}, 0 \)