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Solve Quadratic Equation

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Learn about solving quadratic equations by factoring methods. A quadratic equation is a polynomial equation of degree 2. It can be simply represented in the form: ax2 + bx + c = 0 where 'x' is the constant, a, b, c are the constant co efficient (which are not equal to 0). Method for Solving Quadratic Equations by FactoringBack to Top Here is the method to solving quadratic equations by factoring - In this method, we use the basic idea of factoring the co efficient 'b' in the manner shown:
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Solve Quadratic Equation
Learn about solving quadratic equations by factoring methods.
A quadratic equation is a polynomial equation of degree 2. It can be simply represented
in the form:
ax2 + bx + c = 0
where 'x' is the constant, a, b, c are the constant co efficient (which are not equal to 0).

Method for Solving Quadratic Equations by FactoringBack to Top
Here is the method to solving quadratic equations by factoring -
In this method, we use the basic idea of factoring the co efficient 'b' in the manner
shown:


Know More About What is an Acute Angle

b = a x c
'b' should be broken in such a way that its original value turns out to b equal to the
product of 'a' and 'c'.
Let us take a general description of the method by solving the generalized equation.
Assumption: Let 'p' and 'q' be the constants such that
b = p + q
Also we know the basic requirement for the application of this method, which is
b = a x c
Now to proceed further, we break 'b' such that it fol ows the above mentioned two
conditions. Taking the general equation now :
ax2 + bx + c = 0
ax2 + px + qx +c = 0 (mentioned above: b = p+q)
Now since we have also fol owed the condition b = a x c


Learn More On :- Acute Angle Definition

So 'a' tuns out to be a factor of 'p' and 'q' turns out to be a factor of 'c' also. Moving
forward and taking the common terms out we get :
x(x + p) + q(x + p) = 0
(x + p) x (x + q) = 0
Where 'p' and 'q' are the factors of the equation.
To make the understanding further clear lets take an example
How to Solve Quadratic Equations by Factoring
Below is the example how to solve quadratic equations by factoring method
Let us consider the following quadratic equation :
x2 + x + 6 = 0
here a = 1, b = 1, c = 6.
Now we break 'b = 1' such that the who parts turn out to be the product of 'a x c' i.e.
equal to 6.
x2 + 3x - 2x + 6 = 0 ( both of our necessary conditions are fulfilled here).

Systems Of Equations Solver
Learn about systems of equations concept. A "system" of equations is a set or group
of equations. Linear equations are easy than non-linear equations, and the simplest
linear system is one with two equations and two variables.
System of equations is a col ection of two or additional equations with a same set of
unknowns. In solving a system of equations, we need to find values for every of the
unknowns that will declare every equation in the system. The system of equation can
be linear or non-linear. The problem can be spoken in sequence of actions form or the
problem can be expressed in algebraic form.
Types of Systems of Equations
Elimination :- Elimination technique is considered as one of the algebraic method for
solving systems. In elimination method an operation on 1 equation is performed so that
cancelling one variable and finding the other variable.


Substitution :- In substitution method the algebraic expression of one of the variable is
substituted in another equation at the place of the respective variable and then the

variable is solved. Again by substituting in any of the equation the value of the second
variable is also found.
Linear equation :- An algebraic expression which relate two variables and whose graph
is a line.
Matrix :- A rectangular array of number written in brackets and used to find solutions for
complex systems of equations.
Consistent System :- In this way we have the set of equations whole solutions set is
represented by only one ordered pair.
Solving Systems of Equations
Below are the examples on solving systems of equations
Example 1:
Given :- x + 27 = 71
Solution:
Step 1: x + 27 - 27 = 71 - 21 (Subtract 27 on both the sides)


Read More On :- Define Acute Angle

Step 2: x = 50 (So the answer is 50)
Example 2:
Given
a + 15 = 75.
Solution :
Step 1: We need to find the value of a.
Step 2: Subtract 15 on both the sides.
Step 3: So the value of a is 60.



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