How do use the discriminant to find all values of b for which the equation #2x^2-bx-9=0# has one real root?

1 Answer
Aug 30, 2017

Answer:

See a solution process below:

Explanation:

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 value of #b# where there will be just ONE soluition, we set the discriminate equal to #0#, substitute for #a# and #c# and solve for #b#:

Substitute:

#color(red)(2)# for #color(red)(a)#

#color(blue)(-b)# for #color(blue)(b)#

#color(green)(-9)# for #color(green)(c)#

#color(blue)(-b)^2 - (4 * color(red)(2) * color(green)(-9)) = 0#

#color(blue)(-b)^2 - (-72) = 0#

#color(blue)(-b)^2 + 72 = 0#

#color(blue)(-b)^2 + 72 - color(red)(72) = 0 - color(red)(72)#

#color(blue)(-b)^2 + 0 = -72#

#color(blue)(-b)^2 = -72#

#color(red)(-1) * color(blue)(-b)^2 = color(red)(-1) * -72#

#b^2 = 72#

#sqrt(b^2) = +-sqrt(72)#

#b = +-sqrt(36 * 2)#

#b = +-sqrt(36)sqrt(2)#

#b = +-6sqrt(2)#