Needing help with some algebra. Question says, find all real or imaginary roots, 5x to 2nd power-4x+1=0?

Question

5x^2-4x+1=0
Also says if there is more than one root to separte them with commas (My math class is online) Thank you!!!!!

Answer 1

Use the quadratic formula.
If ax^2 + bx +c = 0, then x = [-b +/- sqrt(b^2 -4ac)]/(2a)
5x^2 - 4x + 1 = 0 means that a = 5, b = -4 and c = 1
Plug in and solve.
x = [-(-4) +/- sqrt((-4)^2 - 4(5)(1))] / (2*5)
x = [4 +/- sqrt(16-20)]/10
x = [4 +/- sqrt(-4)]/10
x = [4 +/- 2i]/10
x = (4+2i)/10 or x = (4-2i)/10
Reduce by dividing everything by 2.
x = (2+i)/5 or x = (2-i)/5

Answer 2

You have ax² + bx + c = 0, where a = 5, b = -4, c = 1.

Checking the discriminant (b² - 4ac) you get (-4)² - 4*5*1 = -4. A negative discriminant means you will have two imaginary roots.

Going further, to solve using the quadratic formula, you should find that you have solutions of:

x = [4 +/- sqrt(-4) ] / 2(5)

That simplifies to 2(2 +/- i)/2(5) and you can cancel a 2.

So your two imaginary roots are:
x = (2 + i)/5, (2 - i)/5

Answer 3

5x^2 - 4x + 1 = 0

Since this is a quadratic equation, you expect there to be two roots.

Solution is easy if the quadratic can be factored, i.e., stated as (x + a)(x + b).

This does not seem to be possible here, so you will have to use the quadratic formula, the general solution for quadratic equations of the form ax^2 + bx + c = 0

x = [- b +- sqrt(b^2 - 4ac)]/2a

The expression b^2 - 4ac is called the discriminant which, if negative, means that the roots are imaginary.

Answer 4

There is a simple formula for equation ax^2+bx+c=0. If the equatino is like this, x will be
x =( (-b)+/-root of (b^2-4ac)/2a
so here,
x = -(-4)+/-root of [(-4)^2-4*5*1] /2*5
=4 +/- (root of -4) /10
=(4+/-(root of (4*(-1))) /10
=(4+/- 2i)/10
as i^2 = -1
so x will be (4+2i)/10 ,(4-2i)/10
x= 2(2+i)/10 , 2(2-i)/10
x= (2+i)/5 , (2-i)/5

Answer 5

I would use the quadratic formula

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

x= [4 +/- (16-4*5*1)^1/2]/2*5

x= [4 +/-(-4)^1/2]/10

x= 4/10 +/- 2i/10

x= 2/5 +/- i/5

which can be written (2+/-i)/5

There are two imaginary roots, [2+i]/5 and [2-i]/5

Answer 6

If you know the quadratic formula, just use that. If not, try to factor the equation :(ax+b)(cx+d) = acx^2 + (ad+bc)x +bd. Thus you need to find a,b,c,d so that ac = 5, ad+bc=-4, bd = 1. Hmm, that looks too hard. Use the quadratic formula.

Answer 7

5x^2 - 4x + 1 = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-4) ± sqrt((-4)^2 - 4(5)(1)))/(2(5))
x = (4 ± sqrt(16 - 20))/10
x = (4 ± sqrt(-4))/10
x = (4 ± 2i)/10
x = (1/5)(2 ± i)

ANS :

x = (1/5)(2 - i) or (1/5)(2 + i)

Answer 8

5x^2-4x+1=0
x=(4+/-sqrt(16-4*5*1))/10=.4+/-sqrt(-4)/10
x=.4+.2i, .4-.2i

Answer 9

5x

Answer 10

uhhhh.... i think it's x=-1
but i'm probably wrong
wait... it's not -1
i don't know i'm only in algebra 1!!!!