Find all real and imaginary solutions of the quadratic equation.?

Question

I am a bit confused on how to do this. When you help me, can you explain how to do it the REAL and IMAGINARY way, please? Thank you

2x^2 - 5x = 0

Answer 1

A lot of quadratics have 3 terms, but this one only has 2. That affects how we'll do it.

2x^2 - 5x = 0

There's an x in both terms, so we'll factor it out to get

x(2x - 5) = 0

Now here's the key to this problem. That thing written just above has two factors, x and 2x-5. When they're multiplied together, they have to equal zero. But as you know, the only way to get zero is for one of the factors to be zero. (Zero times anything is zero.)

On the other hand, we don't care which of the factors is zero, so long as one or the other is. So we set each factor equal to zero, and here's what we get:

(a) x = 0 (b) 2x - 5 = 0

(a) gives us a root right away; when we solve (b), we see that the other root is x = 5/2, or 2.5.

These are the two roots of this equation. Both are real, and there are no imaginary roots. For those, you need a different quadratic.

Hope this lesson helps.

Answer 2

I have no idea whats a real or imaginary solution, but here is how I would do it.

2x^2 - 5x = 0
2xx = 5x
2x = 5
x = 5/2 = 2.5

Answer 3

A quadratic equation
ax^2 + bx + c has roots
x = -(b+(b^2-4ac)^1/2)/2a
and

-(b-(b^2-4ac)^1/2)/2a

The roots of the equation are only imaginary if (b^2-4ac) < 0

In the above equation a = 2, b = -5 and c = 0
therefore roots are
(-(-5)+5)/2*2 and (-(-5)-5)/2*2
=> 5/2,0
This equation has no imaginary roots

Answer 4

taking x common you get x(2x-5)=0
so one of the two multiples must be zero x=0 or 2x-5=0
if x=0 well x=0 but if 2x-5=0 then x=5/2
i dont know if youve noticed but this curve has two real and distinct roots and no imaginary roots

Answer 5

2x^2 - 5x = 0
x(2x - 5) = 0

x = 0 or (5/2)

Answer 6

2x^2-5x-0=0
x^2-5/2x=0
x^2-2*5/4x+25/16-25/16=0
(x-5/4)^2=(5/4)^2
x=0,5/2