3x^2-2x+1=0 solve using the quadratic formula?

(-b+-sqrt(b^2-4ac))/2a

Pug in 3 for a, -2 for b, and 1 for c.

(2+-sqrt(4-12))/6

(2+-sqrt(-8))/6

If you want to reduce the sqrt of -8 by complex numbers, then the answer is (2+-2isqrt(2))/6 or (1+-isqrt(2)/3)

3x^2 - 10x + 5 = 0 a = 3 b = -10 c = 5 [-b ± ?(b^2 - 4ac)] / 2a [-(-10) ± ?(-10)^2 - 4 * 3 * 5)] / (2 * 3) (10 ± ?40) / 6 (10 ± 2?10) / 6 x = (5 ± ?10) / 3 do the others the comparable way

x = [ 4 ± √ (4 - 12 ) ] / 6
x = [ 4 ± √ (8 i² ) ] / 6
x = [ 4 ± (2√2) i ) ] / 6
x = [ 2 ± √2 i ) ] / 3

The equation has no real roots because the discriminant is -8.

No. You can plug the values into the quadratic formula, but you'll find that the discriminant (the number inside the square root sign) is negative, and you can't find any real square root of a negative number.

That said, there will be two complex solutions. If you don't know this, you will if you continue to study maths. All you have to put down is "no real solutions", and you'll get full marks.

3x^2 - 2x + 1 = 0

Compare with general quadratic formula ax^2 + bx + c = 0

Hence a = 3, b = 2 and c = 1

Consider the discriminant in the quadratic formula. That is
(b^2 - 4*a*c)

= 2^2 - (4*3*1)

= 4 - 12

= -8

So the discriminant is negative which means there are no real solutions to the given quadratic equation. If you tried to use the quadratic formula then you have to find the square root of (-8) for which there are no real solutions.

4 + (2i times the sq. root of 2) / 6 or 4 - (2i times the sq. root of two) /6
There are no rational solutions

I just when ahead and work out the math and a solution does not exist.

You can verify my answer by looking at the graph of this function. It never crosses the x-axis. The clossest it gets to crossing it is at the point x=1/3, and y= 2/3, (.333, .666) is the point at the bottem of this u shapped graph.

I hope this helped, have a great night, and good luck with the rest of your math problems. =D

not a real number solution.
Because your quantity under the square root sign in the formula (the b^2 - 4ac part) is going to be negative since
-2^2- 4(3)(1) < 0
and negative square roots are not a real number. They are complex numbers

3x^2 - 2x + 1 = 0
x = [-b ±√(b^2 - 4ac)]/2a

a = 3
b = -2
c = 1

x = [2 ±√(4 - 12)]/6
x = [2 ±√-8]/6 (imaginary number)
(no real roots)