PSAT Math : How to find the solution to a quadratic equation
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Example Question #21 : How To Find The Solution To A Quadratic Equation
The formula to solve a quadratic expression is:
All of the following equations have real solutions EXCEPT:
We can use the quadratic formula to find the solutions to quadratic equations in the form ax2 + 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, b2 – 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 b2 – 4ac. If the value of b2 – 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 b2 – 4ac.
b2 – 4ac = (–4)2 – 4(2)(5) = 16 – 40 = –24 < 0
Because the value of b2 – 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 b2 – 4ac, and thus these other equations will have real solutions.
Example Question #22 : How To Find The Solution To A Quadratic Equation
Let 
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
Add k to both sides.
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.
The answer is 3.
Example Question #21 : Quadratic Equations
Note: Figure NOT drawn to scale.
Refer to the above diagram, which shows Rectangle 
Evaluate 
Insufficient information is given to answer the question.
The corresponding sides of similar triangles are in proportion, so we can set up and solve the proportion statement for 
For the sake of simplicuty, we will let
Since 
Also, 
The proportion statement becomes
Solve for 
By the quadratic equation, setting 
There are two possibilities:
or
Example Question #22 : How To Find The Solution To A Quadratic Equation
Solve for 
Begin by distributing the three on the right side of the equation:
Next combine your like terms by subtracting 
Next, subtract 9 from both sides to give you 
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