Complete the square to identify the two values that solve the following quadratic equation: x2+8x=9

Question

.(1 point)
x=

2. Using the quadratic formula, find the solution to x2+6x+9=0
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)
x=
or
4.
.
How many real solutions does the following quadratic equation have?

25x2+60x+36=0

(1 point)
solution(s)

Answer 1

Let's solve each question step by step:

1. Completing the square for the equation \(x^2 + 8x = 9\):

   To complete the square, follow these steps:

   • Move the constant to the other side: \[ x^2 + 8x - 9 = 0 \]

   • Take half of the coefficient of \(x\) (which is \(8\)), square it, and add it to both sides: \[ \left(\frac{8}{2}\right)^2 = 16 \] Adding \(16\) to both sides gives: \[ x^2 + 8x + 16 = 9 + 16 \]

   • This simplifies to: \[ (x + 4)^2 = 25 \]

   • Now take the square root of both sides: \[ x + 4 = \pm 5 \]

   • Solve for \(x\): \[ x + 4 = 5 \quad \Rightarrow \quad x = 1 \] \[ x + 4 = -5 \quad \Rightarrow \quad x = -9 \]

   So the solutions are: \[ x = 1 \quad \text{or} \quad x = -9 \]

2. Using the quadratic formula for \(x^2 + 6x + 9 = 0\):

   The quadratic formula is given by: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \(a = 1\), \(b = 6\), and \(c = 9\).

   • Calculate the discriminant: \[ b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0 \]

   • Since the discriminant is \(0\), there is one real solution: \[ x = \frac{-6 \pm 0}{2 \cdot 1} = \frac{-6}{2} = -3 \]

   Thus, the solution is: \[ x = -3 \]

3. Solve \(-2x^2 + 3x - 1 = -2\) using the quadratic formula:

   First, rearrange the equation: \[ -2x^2 + 3x - 1 + 2 = 0 \quad \Rightarrow \quad -2x^2 + 3x + 1 = 0 \]

   Multiply through by \(-1\) to make \(a\) positive: \[ 2x^2 - 3x - 1 = 0 \]

   Here, \(a = 2\), \(b = -3\), and \(c = -1\).

   • Calculate the discriminant: \[ b^2 - 4ac = (-3)^2 - 4 \cdot 2 \cdot (-1) = 9 + 8 = 17 \]

   • Using the quadratic formula: \[ x = \frac{-(-3) \pm \sqrt{17}}{2 \cdot 2} = \frac{3 \pm \sqrt{17}}{4} \]

   • We can find the two solutions: \[ x_1 = \frac{3 + \sqrt{17}}{4} \quad \text{(larger)} \] \[ x_2 = \frac{3 - \sqrt{17}}{4} \quad \text{(smaller)} \]

   • Calculate the approximations: \[ \sqrt{17} \approx 4.123 \] \[ x_2 \approx \frac{3 - 4.123}{4} \approx \frac{-1.123}{4} \approx -0.2808 \quad \text{(rounds to -0.28)} \] \[ x_1 \approx \frac{3 + 4.123}{4} \approx \frac{7.123}{4} \approx 1.78075 \quad \text{(rounds to 1.78)} \]

   Thus the solutions are: \[ x = -0.28 \quad \text{or} \quad x = 1.78 \]

4. Number of real solutions for the equation \(25x^2 + 60x + 36 = 0\):

   Use the discriminant to determine the number of solutions: \[ b^2 - 4ac = 60^2 - 4 \cdot 25 \cdot 36 = 3600 - 3600 = 0 \]

   Since the discriminant is \(0\), there is exactly one real solution.

Thus, summarizing the answers:

1. \(x = 1\) or \(x = -9\)
2. \(x = -3\)
3. \(x = -0.28\) or \(x = 1.78\)
4. The number of solution(s): \(1\) real solution.