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EXTRA

Unit 38: Quadratic and Discriminant

The Quadratic Formula and Discriminant

You will need the quadratic formula in three scenarios.

  1. You see something like (5pmsqrt3)  in the answer choices.  The “(pm)” symbol is a clear giveaway that you will have to use the quadratic formula.
  2. You have a quadratic equation that won’t factor, like ({0=x}^2+3x-17)
  3. The question is asking how many solutions there are to a system of equations involving a quadratic (as in the previous unit, but you can’t factor).

Strategy:

  1. Memorize the quadratic formula!!$$x=frac{-bpmsqrt{b^2-4ac}}{2a}$$
  2. Set your quadratic equal to zero and define you’re ab, and c terms. Your a term is the (x^2) coefficient, your b term is the (x) coefficient, and the term is a number.
  3. Plug your terms into the quadratic formula and solve.

Note that sometimes they replace numbers with constants (like p or c).  Don’t let that trip you up!

Also note you can sometimes divide by the (x^2) coefficient to make your life easier – if you have (4x^2+8x-16=0), dividing both sides by 4 gives you (x^2+2x-8) which is easier to work with.

Example:

(x^2+3x+17)

a =1
b = 3
c = 17

(x=frac{-3pmsqrt{3^2-4(1)(17)}}{2(1)})

(x=frac{-3pmsqrt{-57}}2), which breaks into (x=frac{-3+sqrt{-57}}2) and (x=frac{-3-sqrt{-57}}2)

Another example:

$$x^2+px+4q=0$$

If p is equal to 2, what value of q would give this equation one solution?

A) 1
B) 0
C) (frac18)
D) (frac1{32})

Define your a, b, and c terms:

(a=2)
(b=p)
(c=4q) (q is a constant so it has to come along here!)

For this problem, we can use the discriminant. The discriminant is just the stuff under the square root sign in the quadratic formula.

$$b^2-4ac$$

The discriminant is positive2 real solutions
The discriminant is exactly zero1 real solution
The discriminant is negativeNo real solutions

Plug the a, b, and c terms into the discriminant:

$$p^2-4left(2right)left(4qright)$$

$$p^2-32q$$

If p is equal to 2, what value of q would give this equation one solution?

A) 1
B) 0
C) (frac18)
D) (frac1{32})

To have one solution, the discriminant must equal exactly zero.

$$0=4–32q$$

We then solve:

$$32q=4$$

$$q=frac4{32}=frac18$$

So C would be our correct answer.  

If they had instead asked us:

If p is equal to 2, what value of q would give this equation no solutions?

A) 0
B) (frac18)
C) 1
D) -1

Our approach would be the same, except now we’re looking for what would make the discriminant negative: 

$$4-32q<0$$

Our only answer that gives us a negative is C.