Question

How do you find the discriminant of `5x^2-4x+1=3x` and use it to determine if the equation has one, two real or two imaginary roots?

Answer 1

See a solution process below:

Explanation:

First, we need to put the equation in standard form. Subtract `color(red)(3x)` from each side of the equation to put the equation in standard form while keeping the equation balanced:

`5x^2 - 4x - color(red)(3x) + 1 = 3x - color(red)(3x)`

`5x^2 + (-4 - color(red)(3))x + 1 = 0`

`5x^2 + (-7)x + 1 = 0`

`5x^2 - 7x + 1 = 0`

The quadratic formula states:

For `ax^2 + bx + c = 0`, the values of `x` which are the solutions to the equation are given by:

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

The discriminate is the portion of the quadratic equation within the radical: `color(blue)(b)^2 - 4color(red)(a)color(green)(c)`

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

`color(red)(5)` for `color(red)(a)`

`color(blue)(-7)` for `color(blue)(b)`

`color(green)(1)` for `color(green)(c)`

`color(blue)((-7))^2 - (4 * color(red)(5) * color(green)(1))`

`49 - 20`

`29`

Because the discriminate is positive, you will get two real solutions or roots.

Answer 2

See below.

Explanation:

Arrange `5x^2-4x+1=3x`

To get: `5x^2-7x+1=0`

We now have the form:

`ax^2+bx+2`

The discriminant of a quadratic is:

`sqrt(b^2-4ac)`

if: `b^2-4ac > 0` then the roots are real and different.

if: `b^2-4ac =0` then the roots are real and repeated,( this is sometimes just called a single root).

if: `b^2-4ac < 0` then the roots are imaginary ( we will have the root of a negative number).

Hope this helps.