How To Solve A 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 Finding Horizontal Asymptote
b=a×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×c Now to proceed further, we break 'b' such that it follows 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 followed the condition b = a × c Learn More On :- Arc Length of a Circle
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) Ă— (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 FactoringBack to Top 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 Ă— c' i.e. equal to 6. x2 + 3x - 2x + 6 = 0 ( both of our necessary conditions are fulfilled here) Now taking out the common terms gives
How To Solve Absolute Value Equations Learn absolute value equations concept. The equation |x| = 4. This means that x could be 4 or x could be -4. When you take the absolute value of 4, the solution is 4 and when you take the absolute value of -4, the solution is also 4. An absolute value problem, you have to get into account that there can be two solutions that will make the equation true. Learning absolute value equation, you set the quantity inside the absolute value symbol equal to the positive and negative value on the other side of the equal symbol. Solve Absolute Value Equations Below are the examples on how to solve absolute value equations Example 1 :- |x + 1| = 4 Solution :- The quantity inside the absolute value symbol can be equal to 4 or -4
x + 1 = 4 or x + 1 = - 4 Subtract 1 on both side of the given equation x + 1 - 1 = 4 - 1 or x + 1 - 1 = -4 - 1 x = 3 or x = -5 So the solution are x = 3 and x = -5 Example 2: |2x - 3| = x - 5 Solution: When solving this equation, you have to be careful when solving opposite of (x - 5) 2x - 3 = x - 5 or 2x - 3 = -(x - 5) x - 3 = -5 or 2x -3 = -x + 5 x = -2 or 3x – 3 = 5 3x = 8 Read More On :- Finding Arc Length
x= So, the solution is x = -2 and x = 83 Example 3: |x + 1| = 5 Solution: The quantity inside the absolute value symbol can be equal to 5 or -5 x + 1 = 5 or x + 1 = -5 Subtract 1 on both side of the given equation x + 1 - 1 = 5 - 1 or x + 1 - 1 = - 5 - 1 x = 4 or x = - 6 So the solution are x = 4 and x = - 6
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