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Question 1 A) Use the quadratic formula to solve the following quadratic equation: x2+2x−8=0 (1 point) Responses x=−2, x=4 x=−2, x=4 x=2, x=−4 x=2, x=−4 x=10, x=−8 x=10, x=−8 x=−10, x=8 x=−10, x=8 Question 2 A) Use the quadratic formula to solve the following quadratic equation: x2−22x+1=15 (1 point) Responses x=0.6189500386, x=−22.61895004 x=0.6189500386, x=−22.61895004 x=14.74165739, x=7.258342613 x=14.74165739, x=7.258342613 x=−7.258342613, x=−14.74165739 x=−7.258342613, x=−14.74165739 x=22.61895004, x=−0.6189500386 x=22.61895004, x=−0.6189500386 Question 3 A)Using the quadratic formula, find the solution to x2−6x−16=0 .(2 points) x= $$ x= $$ Question 4 A)Which of the following is a solution to the quadratic equation 2x2−5x=6 ? Assume that the solution has been rounded to the nearest hundredth, if applicable.(1 point) Responses x=−1.03 x is equal to negative 1 point 0 3 x=1.03 x is equal to 1 point 0 3 x=−0.89 x is equal to negative 0 point 8 9 x=0.89 x is equal to 0 point 8 9 Question 5 A) How many real solutions does the following quadratic equation have? 4x2+x+3=0 (1 point) Responses no real solutions no real solutions one real solution one real solution three real solutions three real solutions two real solutions

1 answer

Let's solve each question step-by-step.

Question 1

To solve the quadratic equation \(x^2 + 2x - 8 = 0\) using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):

  • Here, \(a = 1\), \(b = 2\), and \(c = -8\).
  • First, calculate the discriminant: \(b^2 - 4ac = 2^2 - 4(1)(-8) = 4 + 32 = 36\).
  • Now apply the quadratic formula: \[ x = \frac{-2 \pm \sqrt{36}}{2(1)} = \frac{-2 \pm 6}{2} \]
  • This results in: \[ x = \frac{4}{2} = 2 \quad \text{and} \quad x = \frac{-8}{2} = -4 \]

So, the solutions are \(x = 2\) and \(x = -4\). Response: x=2, x=−4

Question 2

To solve the quadratic equation \(x^2 - 22x + 1 = 15\), first rearrange it to standard form: \[ x^2 - 22x - 14 = 0 \] Now apply the quadratic formula with \(a = 1\), \(b = -22\), and \(c = -14\):

  • The discriminant is: \[ b^2 - 4ac = (-22)^2 - 4(1)(-14) = 484 + 56 = 540 \]

  • Apply the formula: \[ x = \frac{22 \pm \sqrt{540}}{2} \] Simplifying \(\sqrt{540} = \sqrt{36 \cdot 15} = 6\sqrt{15}\): \[ x = \frac{22 \pm 6\sqrt{15}}{2} = 11 \pm 3\sqrt{15} \] Calculating the approximations:

    \(x \approx 11 + 3(3.87298) \approx 11 + 11.61894 \approx 22.61894\) and \(x \approx 11 - 11.61894 \approx -0.61894\).

Thus, the solutions are approximately \(x \approx 22.61895\) and \(x \approx -0.61895\). Response: x=22.61895004, x=−0.6189500386

Question 3

To solve \(x^2 - 6x - 16 = 0\) using the quadratic formula with \(a = 1\), \(b = -6\), and \(c = -16\):

  • The discriminant is: \[ (-6)^2 - 4(1)(-16) = 36 + 64 = 100 \]
  • Applying the quadratic formula: \[ x = \frac{-(-6) \pm \sqrt{100}}{2(1)} = \frac{6 \pm 10}{2} \]
  • This results in: \[ x = \frac{16}{2} = 8 \quad \text{and} \quad x = \frac{-4}{2} = -2 \]

Responses: x=8, x=−2

Question 4

The equation \(2x^2 - 5x = 6\) rearranged becomes \(2x^2 - 5x - 6 = 0\). Now set \(a = 2\), \(b = -5\), and \(c = -6\):

  • Discriminant: \[ (-5)^2 - 4(2)(-6) = 25 + 48 = 73 \]
  • Using the quadratic formula: \[ x = \frac{-(-5) \pm \sqrt{73}}{2(2)} = \frac{5 \pm \sqrt{73}}{4} \]
  • Approximating \(\sqrt{73} \approx 8.544\): \[ x \approx \frac{5 + 8.544}{4} \approx 3.886 \quad \text{and} \quad x \approx \frac{5 - 8.544}{4} \approx -0.886 \]

Thus, rounding to the nearest hundredth gives approximately \(x \approx 0.89\) and \(x \approx -1.03\). Response: x=−1.03

Question 5

For the quadratic equation \(4x^2 + x + 3 = 0\):

  • The discriminant: \[ 1^2 - 4(4)(3) = 1 - 48 = -47 \] Since the discriminant is negative, there are no real solutions. Response: no real solutions