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?

2 Answers
Oct 2, 2017

Answer:

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.

Oct 2, 2017

Answer:

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.