SOLUTION: I am trying to find b^2-4ac and the number of real solutions for the equation 16-24x+9x^2=0. If I put it in standard form ax^2 + ax + a = 0 a=-9 b=24 c=-16 b^2-4ac= 24^2

Question 151472: I am trying to find b^2-4ac and the number of real solutions for the equation 16-24x+9x^2=0.

If I put it in standard form ax^2 + ax + a = 0
a=-9 b=24 c=-16
b^2-4ac= 24^2 - 4(-9)(-16)=576-576= 0
So: when the discriminat equals 0 there should be only one solution correct? But how do I find that 1 solution?
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
Start with the given equation.

Rearrange the terms.

From we can see that , , and

Start with the discriminant formula.

Plug in , , and

Square to get

Multiply to get

Subtract from to get

Since the discriminant is equal to zero, this means that there is one real solution.

So you are correct. However, I'm not sure where you got a=-9 b=24 c=-16 from.

---------------------------------------

To find the one solution, you have 2 options

Option # 1 Quadratic Formula (preferred method)

Start with the given equation.

Notice we have a quadratic equation in the form of where , , and

Let's use the quadratic formula to solve for x

Start with the quadratic formula

Plug in , , and

Negate to get .

Square to get .

Multiply to get

Subtract from to get

Multiply and to get .

Take the square root of to get .

or Break up the expression.

or Combine like terms.

or Simplify.

So our answer is (with a multiplicity of 2)

Option # 2 Factoring:

Start with the given equation

Factor the left side (note: if you need help with factoring, check out this solver)

Now set each factor equal to zero:
or

or Now solve for x in each case

Since we have a repeating answer, our only answer is

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