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Using Quadratic Formulas in Real Life Situations
This lesson will show us how to use the quadratic formula in real-life situations. We'll look at a couple of different real-world examples of when this formula can be used to solve problems that can be modeled using quadratic equations.

SOHCAHTOA Examples Formula Rules
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Relation in Math: Definition & Examples
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Radical Expression: Definition & Examples
Mathematical expressions that contain a radical symbol (√) are called radical expressions. Learn how to define radical expressions, the history of the term radical, how to use examples to solve radical equations, and special circumstances when radical expressions have no real roots.
Quadratic equations
Step by step solution :
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring x2-4x-2
The first term is, x2 its coefficient is 1 .
The middle term is, -4x its coefficient is -4 .
The last term, "the constant", is -2
Step-1 : Multiply the coefficient of the first term by the constant 1 • -2 = -2
Step-2 : Find two factors of -2 whose sum equals the coefficient of the middle term, which is -4 .
-2 | + | 1 | = | -1 | |
-1 | + | 2 | = | 1 |
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 1 :
x2 - 4x - 2 = 0Step 2 :
Parabola, Finding the Vertex :
2.1 Find the Vertex of y = x2-4x-2
Parabolas have a highest or a lowest point called the Vertex . Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) . We know this even before plotting "y" because the coefficient of the first term, 1 , is positive (greater than zero).
Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.
Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is 2.0000
Plugging into the parabola formula 2.0000 for x we can calculate the y -coordinate :
y = 1.0 * 2.00 * 2.00 - 4.0 * 2.00 - 2.0
or y = -6.000
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = x2-4x-2
Axis of Symmetry (dashed) {x}={ 2.00}
Vertex at {x,y} = { 2.00,-6.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-0.45, 0.00}
Root 2 at {x,y} = { 4.45, 0.00}
Solve Quadratic Equation by Completing The Square
2.2 Solving x2-4x-2 = 0 by Completing The Square .
Add 2 to both side of the equation :
x2-4x = 2
Now the clever bit: Take the coefficient of x , which is 4 , divide by two, giving 2 , and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have :
2 + 4 or, (2/1)+(4/1)
The common denominator of the two fractions is 1 Adding (2/1)+(4/1) gives 6/1
So adding to both sides we finally get :
x2-4x+4 = 6
Adding 4 has completed the left hand side into a perfect square :
x2-4x+4 =
(x-2) • (x-2) =
(x-2)2
Things which are equal to the same thing are also equal to one another. Since
x2-4x+4 = 6 and
x2-4x+4 = (x-2)2
then, according to the law of transitivity,
(x-2)2 = 6
We'll refer to this Equation as Eq. #2.2.1
The Square Root Principle says that When two things are equal, their square roots are equal.
Note that the square root of
(x-2)2 is
(x-2)2/2 =
(x-2)1 =
x-2
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-2 = √ 6
Add 2 to both sides to obtain:
x = 2 + √ 6
Since a square root has two values, one positive and the other negative
x2 - 4x - 2 = 0
has two solutions:
x = 2 + √ 6
or
x = 2 - √ 6
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving x2-4x-2 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = 1
B = -4
C = -2
Accordingly, B2 - 4AC =
16 - (-8) =
24
Applying the quadratic formula :
4 ± √ 24
x = —————
2
Can √ 24 be simplified ?
Yes! The prime factorization of 24 is
2•2•2•3
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 24 = √ 2•2•2•3 =
± 2 • √ 6
√ 6 , rounded to 4 decimal digits, is 2.4495
So now we are looking at:
x = ( 4 ± 2 • 2.449 ) / 2
Two real solutions:
x =(4+√24)/2=2+√ 6 = 4.449
or:
x =(4-√24)/2=2-√ 6 = -0.449
Two solutions were found :
- x =(4-√24)/2=2-√ 6 = -0.449
- x =(4+√24)/2=2+√ 6 = 4.449

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The solution is 0<x<4.
But I think the solution should be x<0 x>4.
Can someone please assist? Thanks.

The solution is 0<x<4.
But I think the solution should be x<0 x>4.
Can someone please assist? Thanks.
Hi happyapple123,
\(x^2 - 4x <0 => x (x - 4) < 0\)
This means either \(x < 0\) or \((x - 4) < 0\) but not both, because if both are negative then multiplication of two negative numbers is positive so the equation will not be true.
Now, if
\(x < 0\) then \((x - 4)\) is also less than 0, which makes both negative, so this cannot be true.
for example,
if \(x = -1\), then \((x - 4) => (-1 - 4) = -5 < 0\)
=> so, \((-1) * (-5) = 5 > 0\), so equation doesn't agree.
Hence, only \((x - 4) < 0\) but \(x > 0\) which means
\(0 < x < 4\)

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Linear equations with one unknown
Step by step solution :
Step 1 :
Step 2 :
Pulling out like terms :
2.1 Pull out like factors :
x2 - 4x = x • (x - 4)
Equation at the end of step 2 :
x • (x - 4) = 0Step 3 :
Theory - Roots of a product :
3.1 A product of several terms equals zero.
When a product of two or more terms equals zero, then at least one of the terms must be zero.
We shall now solve each term = 0 separately
In other words, we are going to solve as many equations as there are terms in the product
Any solution of term = 0 solves product = 0 as well.
Solving a Single Variable Equation :
3.2 Solve : x = 0
Solution is x = 0
Solving a Single Variable Equation :
3.3 Solve : x-4 = 0
Add 4 to both sides of the equation :
x = 4
Two solutions were found :
- x = 4
- x = 0
2 4x solve
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and then it came.
Completing The Square Method and Solving Quadratic Equations - Algebra 2I also said the same thing and bought several bottles of wine for my wife and cognac for us men. Having laid out the bought, building in our house, edibles in the house of Matveich in the evening they put together a table. Taking home things with us, we changed into T-shirt shorts and sat down at the table.
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