x2+6x+9=0

Question

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.
and

Answer 1

Let's solve each equation step by step.

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

This is a perfect square trinomial: \[ (x + 3)^2 = 0 \] Thus, the solution is: \[ x = -3 \]

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

This is a difference of squares: \[ (4x)^2 - 7^2 = 0 \] Factoring gives us: \[ (4x - 7)(4x + 7) = 0 \] Setting each factor to zero:

1. \( 4x - 7 = 0 \) → \( 4x = 7 \) → \( x = \frac{7}{4} \)
2. \( 4x + 7 = 0 \) → \( 4x = -7 \) → \( x = -\frac{7}{4} \)

The solutions in fraction form, listed from smaller to larger, are: \[ x = -\frac{7}{4} \quad \text{and} \quad x = \frac{7}{4} \]

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

We can use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 3, b = 18, c = 24 \).

Calculate the discriminant: \[ b^2 - 4ac = 18^2 - 4(3)(24) = 324 - 288 = 36 \] Now substituting back into the quadratic formula: \[ x = \frac{-18 \pm \sqrt{36}}{2(3)} = \frac{-18 \pm 6}{6} \] Calculate the two solutions:

1. \( x = \frac{-18 + 6}{6} = \frac{-12}{6} = -2 \)
2. \( x = \frac{-18 - 6}{6} = \frac{-24}{6} = -4 \)

The solutions, listed from smaller to larger, are: \[ x = -4 \quad \text{and} \quad x = -2 \]

4. Solve \( 2x^2 + x - 5 = 0 \)

Using the quadratic formula again: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 2, b = 1, c = -5 \] Calculate the discriminant: \[ b^2 - 4ac = 1^2 - 4(2)(-5) = 1 + 40 = 41 \] Substitute into the formula: \[ x = \frac{-1 \pm \sqrt{41}}{4} \] Calculating both solutions:

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

Approximating the second solution (using \( \sqrt{41} \approx 6.4 \)): \[ x \approx \frac{-1 - 6.4}{4} \approx \frac{-7.4}{4} \approx -1.85 \]

Now let’s calculate the first solution: \[ x \approx \frac{-1 + 6.4}{4} \approx \frac{5.4}{4} \approx 1.35 \]

The solutions, rounded to the hundredths and listed from smaller to larger, are: \[ x \approx -1.85 \quad \text{and} \quad x \approx 1.35 \]

5. Find the zeros of \( y = 2x^2 + 7x \)

To find the zeros, set \( y = 0 \): \[ 2x^2 + 7x = 0 \] Factoring out \( x \): \[ x(2x + 7) = 0 \] Setting each factor to zero:

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

The zeros, listed from smaller to larger, are: \[ x = -\frac{7}{2} \quad \text{and} \quad x = 0 \]

Summary of Solutions:

1. \( x = -3 \)
2. \( x = -\frac{7}{4} \) and \( x = \frac{7}{4} \)
3. \( x = -4 \) and \( x = -2 \)
4. \( x \approx -1.85 \) and \( x \approx 1.35 \)
5. \( x = -\frac{7}{2} \) and \( x = 0 \)