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Quadratic Equations

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A Quadratic Equation is an polynomial equation with highest degree 2.

 For e.g.:

 5x2 + 2=0
2x2 + x + 1=0

 General form of the equation is:

 ax2 + bx + cx = 0 where a is not equal to 0

 Normal form of Quadratic equation can also be represented by two parameters. (say m and n)

x2 + mx + n = 0

 Normally , We are asked to find the roots of the equation. The roots of the equation can be found using three methods.

 Let us take the example x2 -6x + 8=0 and solve this equation using these methods separately. Let's Start:

 1.) The Grouping Method or that's what i call it

    In this method we try to split the middle term or 'b' such that it is equal to the product of the rest of the terms or 'ac'.

 Now, let us apply this method on the above equation.

 x2 -6x + 8=0

 As, it is easily visible, the term -6 has to be split in a such a way that it's product equals to 8.
x2 -6x + 8=0

 So, -6 can be written as -4 - 2 and there product is also equal to 8.

 Now, the equation is something like this:

 x2 -4x - 2x + 8=0

 taking common 

 x(x-4) -2(x-4)=0
Again, taking common,

 (x-2)(x-4)=0

 As we can see the roots are simply 2 and 4. I advise you to use this method when dealing with composite integers which are also small.

Also, this method is only applicable when the equation has rational roots and most of the times integral roots.

 2.) Completing square method

 In this method we try to convert the variable part or more precisely 
the ax2 + bx part into a perfect square.

 For doing so, we should be perfectly aware of this identity i.e.

 (d+e)2= d2 + 2de + e2 

 So, all we have to do is to create a perfect square. Let's take the example above, so

 x2 -6x + 8=0

 Now, we can see that according to the identity all we have to do is to divide the coefficient of x by 2 and the coefficient left is the required value of e.

 So, x2 -2*3x + (3)2 - (3)2 + 8=0

 So, it becomes (x-3)2 + 8 -9=0
 i.e

 (x-3)2  = 1

 i.e. either (x-3) = -1 or 1

 So, x is either 2 or 4

 Hence, we found the roots of this equation.

 Now, if we extend this method to general form of equation i.e. 

 ax2 + bx + cx = 0 we would be able to come up with a general formula 
and this formula is actually our third way to solve for the roots of quadratic equation.
So, let's start solving for the equation...

 ax2 + bx + c = 0

 dividing and multiplying b by 2 gives...

 ax2 + 2*(b/2)x + c = 0

 Now, we can see that we have to add (b/2)2 both sides giving us,
ax2 + 2*(b/2)x + (b/2)2 + c = 0 + (b/2)2  

 Now, completing the square gives,

 (ax + b/2)2 = b2/4 - c

 Now, by simplifying the equation gives,

Solving this gives the famous formula of all times,

 x= ( -b ±√ (b2 - 4ac) )/2a

 3.) Using the above formula

 So, again using the same equation x2 -6x + 8=0 gives,

 x= ( -(-6) ±√ (62 - 4*1*8) )/2*1

 On solving, it yields the same results as,

 x=2 and x=4.

 Also, we can infer from the above equation that 'a' is in the denominator and thus a ≠ 0.

 This was all for the Board exams, I will be writing more about Quadratic equations and How to solve them using parabolas. So, guys keep tuning in. There's a video coming for the same.
Don't forget to share and like.

 For Doubts and Correction kindly comment below :)

 Thank you









  

Comments

  1. Unknown5 September 2016 at 06:59

    If you want summary on other topics kindly comment and you might be responsible for the next post. Thanks!

    ReplyDelete

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