Question

Find the number of​ real-number solutions of the equation below. x^2-3x+9=0

Answer 1

We can use the quadratic formula to solve for the roots of the equation:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = -3, and c = 9. Plugging in these values, we get:

x = (-(-3) ± sqrt((-3)^2 - 4(1)(9))) / 2(1)
x = (3 ± sqrt(-27)) / 2

Since the square root of a negative number is not a real number, there are no real solutions to this equation.

Answer 2

Find the number of​ real-number solutions of the equation below. x^2+5x-8=0

Answer 3

We can use the quadratic formula to solve for the roots of the equation:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

In this case, a = 1, b = 5, and c = -8. Plugging in these values, we get:

x = (-5 ± sqrt(5^2 - 4(1)(-8))) / 2(1)
x = (-5 ± sqrt(89)) / 2

Since the square root of 89 is a positive real number, there are two real solutions to this equation.

Answer 4

Find the number of​ real-number solutions of the equation below. 9x^2+18x+9=0

Answer 5

We can start by factoring out the common factor of 9:

9x^2 + 18x + 9 = 9(x^2 + 2x + 1)

Then we can recognize that the expression inside the parentheses is a perfect square trinomial:

9x^2 + 18x + 9 = 9(x + 1)^2

Now we have factored the equation into a product of two factors, and we can see that the only real solution occurs when the square factor equals zero:

(x + 1)^2 = 0

Solving for x, we get:

x + 1 = 0

x = -1

So there is only one real solution to this equation: x = -1.

Answer 6

It asked for the number of solutions, so all we had to do is
evaluate the discriminant b^2 - 4ac

there was no need to actually find the roots like the bot did

Answer 7

You are correct. The number of real solutions of a quadratic equation can be determined by evaluating the discriminant b^2 - 4ac.

In the case of 9x^2 + 18x + 9 = 0, we have a = 9, b = 18, and c = 9.

The discriminant is b^2 - 4ac = 18^2 - 4(9)(9) = 0

Since the discriminant is zero, we know that there is only one real solution to this equation.