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39 votes
39 votes
Solve the quadratic equation

Solve the quadratic equation-example-1
User Kreig
by
2.1k points

2 Answers

15 votes
15 votes

Answer:

Our problem is
x^2-2x+6=0, but as we can see, we are unable to factor. We have to use the quadratic equation to solve.


x^2-2x+6=0


(-b+-√(b^2-4ac))/(2a)

Positive Quadratic Formula:


(-(-2)+√((-2)^2-4(1)(6)))/(2(1))


=(2+√(4-24))/(2)\\=1+(√(-20))/(2)

Negative Quadratic Formula:


(-(-2)-√((-2)^2-4(1)(6)))/(2(1))


=(2-√(4-24))/(2)\\=1-(√(-20))/(2)

Since both of our answers are the square root of a negative number, we know that the quadratic equation has no real solution.

*We could have also used the Discriminant Test to determine whether the quadratic equation has real roots or not. However, for our means, the quadratic equation seems enough.

User Burhanuddin Rashid
by
2.7k points
17 votes
17 votes

Answer:

D. No real Solution

Explanation:

Hello!

Let's use the quadratic formula:
x = (-b \pm √(b^2 - 4ac))/(2a)

In our equation:

  • a = 1
  • b = -2
  • c = 6

This comes from the standard form of a quadratic:
ax^2 + bx + c

Now, solve:


  • x = (-b \pm √(b^2 - 4ac))/(2a)

  • x = (-(-2) \pm √((-2)^2 - 4(1)(6)))/(2(1))

  • x = (2 \pm √(4 -24))/(2)

  • x = (2 \pm√(-20))/(2)

  • x = (2 \pm 2√(-5))/(2)

  • x = 1 \pm \sqrt{-5

Since the radicand (-5) is negative, there are no real solutions. The correct answer is Option D.

User Tupshin Harper
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3.1k points