Equation help

Question

Consider the equation 9x2 + 12x - 5=0.

(a) Find and state the value of the discriminant, b2 – 4ac. Then state whether one real-number solution, two different real-number solutions, or two different imaginary-number solutions exists. Show work.

 

 

(b) Find the exact solutions of the equation. Simplify as much as possible. Show work. 

Answer 1

Hello, thank you for taking the time to post your question!

The given quadratic equation is

9x^2 + 12x – 5 = 0, meaning that a = 9, b = 12, c = -5 for this expression

For the discriminant you can just plug in those values directly

Discriminant = 12^2 – 4(9)(-5) = 144 + 180 = 324

Since the value exceeds 0 that means that there are two different real-number solutions on this one

To find those values then for part (b) I think that it makes the most sense to use the quadratic formula or some type of technology since this one doesn’t factor neatly. When I do that I get x = 1/3 and x = -5/3 for the exact values for this quadratic!

I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)

Answer 2

Recall that the solution to a quadratic equation of the form: ax²+bx+c=0 is given by:

 

   x=[-b±√(b²-4ac)]/2a

 

Note that the discriminant is the value under the square root, hence is:

 

i) A real number if b²-4ac>0 (because we can find the square root of a positive number)

ii) Zero if the number b²-4ac=0  (because the square root of zero is zero)

iii) A complex number if b²-4ac<0  (because the square root of a negative number has a factor of i)

 

Due to the ± preceding the square root in the quadratic equation, cases i) and iii) above will correspond to two distinct values for x

 

a) Given 9x²+12x-5-0, we have that:

 

   a=9

   b=12

   c=-5

 

therefore the discriminant b²-4ac = 12²-4(9)(-5)

                                              = 144+180

                                              = 324

                                              > 0   →  two distinct real roots

 

 

b) x=[-b±√(b²-4ac)]/2a

 

     =[-12±√(12²+12(15))]/2(9)

 

     =[-12±√[12(12+15)]]/18

 

     =[-12±√[12(27)]]/18

 

     =[-12±√(4(3⁴))]/18

 

     =[-12±2(3²)]/18

 

     =[-12±18]/18

 

     ={(-12+18)/18, (-12-18)/18}

 

     ={6/18, -30,18}

 

     ={1/3, -5/3}