is x^2-2x+4 even possible?

Question

I used the quadratic equation and keep coming up with a negative radicand! It's asking me to find the x-intercepts if y=0, and the equation is y = x^2 – 2x + 4

Answer 1

Since the last term is positive 4 that means that the curve's minimum point is at y=4. So, it never intercepts the x - axis. So the answer in undefined.

Answer 2

Unfortunately this equation does not cross the x-axis as you have already discovered. You don't need to do the complete quadratic to check just the bit under the square root:

b^2 - 4ac

= (-2 ^2) - (4 * 1 * 4)

= 4 - 16 = -12

A negative here means it wont cross the x-axis.

Answer 3

The discriminant of ax^2 + bx + c = 0 is b^2 - 4ac.
For your equation this is:
2^2 - 4*4 < 0.
The equation therefore has no real roots, and the graph does meet the x axis.

Answer 4

It's impossible as D=4-16 < 0

Answer 5

y=x^2-2x+4=0
discriminant=2-16=-12 negative
s o the equation does not have real roots,so this parabola does not have intercection points with the x-axis

Answer 6

This quadratic equation does not have any real roots: that is, the parabola never crosses the x-axis. The parabola is always above the x-axis.

The roots are said to be imaginary as they involve i =sqrt(-1), or, as some like to say i^2= -1.

Answer 7

2 +/- √4 (-4)(1)(4) / (2)

= 2+/- √(4 * -16) / (2)

= (2 +/- √(-64))/ (2)

= 2 +/- 8i /(2)

NO REAL NUMBER SOLUTIONS

Answer 8

nope

Answer 9

the answer is no answer cause you cannot simplify it