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Quadratic Equation

What is a Quadratic Equation

An equation of the form ax2 + bx + c = 0 where a≠0 is called a quadratic equation.

Features of a quadratic equation

  1. It consists of one variable.
    The variable in the above case is x.

  2. The variable is raised to different powers.
    The powers to which it is raised is 0, 1 and 2.
    The result is x0 or 1, x1 or x and x2.

  3. The variables raised to different powers are multiplied by some constants.a,b and c.
    c×1,bx and ax2.

  4. Each term obtained is added.
    ax2 + bx + c.

  5. The highest degree of the variable is 2 with coefficient not equal to zero i.e a≠0.

  6. The whole expression is equated to zero.
    ax2 + bx + c = 0

Root of a quadratic Equation

A value which satisfies the quadratic equation is called the root of the quadratic equation. If α is a root of ax2 + bx + c = 0 then aα2 + bα + c = 0.
There are two roots of a quadratic equation.

Solving a quadratic equation by factorization

  1. Reduce the equation to the quadratic form i.e. in the form ax2 + bx + c = 0.
  2. Factorize the expression on the left hand side.
  3. Put each factor to zero and solve.

Solving a quadratic equation by formula

The roots of a quadratic equation can be obtained by using the formula
solution of a quadratic equation

Solutions

The equation is ax2 + bx + c = 0.
It can be written as a(x2 + (b/a)x + (c/a)) = 0
As a≠0 because if it is zero then the equation is not quadratic.
So, (x2 + (b/a)x + (c/a)) = 0

Solution 1

A quadratic equation can be solved by removing the second term in ax2 + bx + c = 0. This can be achieved by completing the square.
(x2 + (b/a)x + (c/a)) = 0
Completing the square
(x2 + (2b/2a)x + b2/4a2 - b2/4a2 + (c/a)) = 0
(x2 + (2b/2a)x + b2/4a2) = b2/4a2 - (c/a)
(x + (b/2a))2 = b2/4a2 - (c/a)
(x + (b/2a))2 = (b2 - 4ac)/4a2
(x + (b/2a)) = ±√[(b2 - 4ac)/4a2]
x = -(b/2a) ±√[(b2 - 4ac)/4a2]
x = [-b ± √(b2 - 4ac)]/2a


The two roots are
x = [-b + √(b2 - 4ac)]/2a
x = [-b − √(b2 - 4ac)]/2a

Solution 2

(x2 + (b/a)x + (c/a)) = 0
We can write it as
x[x + (b/a)] = -(c/a)
Let A = x and B = x + b/a then
B - A = b/a , B + A = 2x + b/a and AB = -c/a
As we know the identity
(B + A)2 = (B - A)2 + 4AB
(2x + b/a)2 = (b/a)2 - 4c/a
2x + b/a = ± √[(b2 - 4ac)/a2]
2x = (-b/a) ± √[(b2 - 4ac)/a2]
x = [-b ± √(b2 - 4ac)]/2a


The two roots are
x = [-b + √(b2 - 4ac)]/2a
x = [-b − √(b2 - 4ac)]/2a

Solution 3

(x2 + (b/a)x + (c/a)) = 0
As (x - A)(x - B) = x2 - (A+B)x + AB
Here A and B are roots
Comparing this with equation we get
B + A = -b/a and AB = c/a
As we know the identity
(B - A)2 = (B + A)2 - 4AB
(B - A)2 = (-b/a)2 - 4c/a
(B - A)2 = (b2 - 4ac)/a2
(B - A) = + √[(b2 - 4ac)/a2]
B + A + B - A = 2B = (-b/a) + √[(b2 - 4ac)/a2]
B = (-b/2a) + √(b2 - 4ac)/2a
B = [-b + √(b2 - 4ac)]/2a
A = B + A - B = (-b/2a) - √[(b2 - 4ac)]/2a
A = [-b - √(b2 - 4ac)]/2a


The two roots are
x = [-b + √(b2 - 4ac)]/2a
x = [-b − √(b2 - 4ac)]/2a
From solution 3 we find that if we are given roots then we can frame a quadratic equation.
Suppose A and B are roots of a quadratic equation then the quadratic equation is x2 -(A+B)x + AB = 0.
For example,
4 and 5 are roots of the equation x2 -(4+5)x + 4×5 = 0
x2 - 9x + 20 = 0



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