# PSAT Math : How to find the solution to a quadratic equation

## Example Questions

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### Example Question #331 : Algebra

The formula to solve a quadratic expression is:

All of the following equations have real solutions EXCEPT:

Explanation:

We can use the quadratic formula to find the solutions to quadratic equations in the form ax+ bx + c = 0. The quadratic formula is given below.

In order for the formula to give us real solutions, the value under the square root, b– 4ac, must be greater than or equal to zero. Otherwise, the formula will require us to find the square root of a negative number, which gives an imaginary (non-real) result.

In other words, we need to look at each equation and determine the value of b 4ac. If the value of b– 4ac is negative, then this equation will not have real solutions.

Let's look at the equation 2x2 – 4x + 5 = 0 and determine the value of b– 4ac.

b– 4ac = (–4)2 – 4(2)(5) = 16 – 40 = –24 < 0

Because the value of b– 4ac is less than zero, this equation will not have real solutions. Our answer will be 2x2 – 4x + 5 = 0.

If we inspect all of the other answer choices, we will find positive values for b– 4ac, and thus these other equations will have real solutions.

### Example Question #302 : Equations / Inequalities

Let , and let . What is the sum of the possible values of such that .

Explanation:

We are told that f(x) = x2 - 4x + 2, and g(x) = 6 - x. Let's find expressions for f(k) and g(k).

f(k) = k2 – 4k + 2

g(k) = 6 – k

Now, we can set these expressions equal.

f(k) = g(k)

k2 – 4k +2 = 6 – k

k2 – 3k + 2 = 6

Then subtract 6 from both sides.

k2 – 3k – 4 = 0

Factor the quadratic equation. We must think of two numbers that multiply to give us -4 and that add to give us -3. These two numbers are –4 and 1.

(k – 4)(k + 1) = 0

Now we set each factor equal to 0 and solve for k.

k – 4 = 0

k = 4

k + 1 = 0

k = –1

The two possible values of k are -1 and 4. The question asks us to find their sum, which is 3.

### Example Question #161 : Equations / Inequalities

Note: Figure NOT drawn to scale.

Refer to the above diagram, which shows Rectangle  with .

is the midpoint of

Evaluate  (to the nearest tenth, if applicable).

Insufficient information is given to answer the question.

Explanation:

The corresponding sides of similar triangles are in proportion, so we can set up and solve the proportion statement for :

, so

For the sake of simplicuty, we will let

Since  is the midpoint of .

Also, .

The proportion statement becomes

Solve for  using cross-products:

By the quadratic equation, setting :

There are two possibilities:

or

is divided into segments of length 2.9 and 17.1. The lesser is the length of , so the correct choice is 2.9.

Solve for