**The Quadratic Formula and Discriminant**

You will need the quadratic formula in three scenarios.

- 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.
- You have a quadratic equation that won’t factor, like ({0=x}^2+3x-17)
- 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****:**

**Memorize the quadratic formula!!**$$x=frac{-bpmsqrt{b^2-4ac}}{2a}$$**Set your quadratic equal to zero and define you’re***a*,*b*, and*c*terms. Your*a*term is the**(x^2) coefficient, your***b*term is the (x) coefficient, and the*c*term is a number.**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 positive | 2 real solutions |

The discriminant is exactly zero | 1 real solution |

The discriminant is negative | No 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.

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