Miraculous world of Numbers: 60-01 Basics of Quadratic Equations

Monday, May 20, 2013

60-01 Basics of Quadratic Equations

An equation of the form a x ² + b x + c = 0 where a, b, and c are the real numbers and a ≠ 0 is known as a quadratic equation. Here "x" is the variable. The highest power of variable x is 2 so it is an equation of order 2. That's why it is known as a quadratic equation. Here "a" is the coefficient of x ², "b" is the coefficient of x, and "c" is the constant term of the quadratic equation.

Roots of the quadratic equation:

The values of the variable "x" which satisfy the equation are known as the roots of the equation. For the quadratic equation, we have two roots. To understand the concept of the roots, let's see the following case which is taken in the reverse direction.

Let us take the roots of the quadratic equation as x = 2 and x = 3.

x = 2 so (x - 2) = 0

x = 3 so (x - 3) = 0.

so taking the product of these two brackets we have,

(x - 2) (x - 3) = 0

x ² - 2 x - 3 x + 6 = 0

x ² - 5 x + 6 = 0

So, x ² - 5 x + 6 = 0 is the quadratic equation in "x" which has two roots 2 and 3.

Now we will study a few examples of roots of the quadratic equation.

1] Check whether the values given against the quadratic equation are the roots of the equation or not.

a] 3 x ² + 7 x + 2 = 0, -1, -2, -1/3, and -3.

Solution:

1) Let us take, L H S = 3 x ² + 7 x + 2

2) We need to check whether -1, -2, -1/3, and -3 are the roots of 3 x ² + 7 x + 2 = 0 or not,

3) so put x = - 1 in L H S, we get,

L H S = 3 x ² + 7 x + 2
L H S = 3 (-1) ² + 7 (-1) + 2
L H S = 3 (1) - 7 + 2
L H S = 3 - 7 + 2
L H S = 5 - 7
L H S = - 2 ≠ 0
L H S ≠ R H S.
So, -1 is not a root of the quadratic equation 3 x ² + 7 x + 2 = 0.

4) so put x = - 2 in L H S, we get,

L H S = 3 x ² + 7 x + 2
L H S = 3 (-2) ² + 7 (-2) + 2
L H S = 3 (4) - 14 + 2
L H S = 12 - 14 + 2
L H S = 14 - 14
L H S = 0
L H S = R H S.
So, -2 is a root of the quadratic equation 3 x ² + 7 x + 2 = 0.

5) so put x = - 1/3 in L H S, we get,

L H S = 3 x ² + 7 x + 2
L H S = 3 (-1/3) ² + 7 (-1/3) + 2
L H S = 3 (1/9) - 7/3 + 2
L H S = 3/9 - 7/3 + 2
L H S = 1/3 - 7/3 + 2
L H S = (1 - 7)/3 + 2
L H S = - 6/3 + 2
L H S = - 2 + 2
L H S = 0
L H S = R H S.
So, -1/3 is a root of the quadratic equation 3 x ² + 7 x + 2 = 0.

6) so put x = - 3 in L H S, we get,

L H S = 3 x ² + 7 x + 2
L H S = 3 (-3) ² + 7 (-3) + 2
L H S = 3 (9) - 21 + 2
L H S = 27 - 21 + 2
L H S = 6 + 2
L H S = 8 ≠ 0
L H S ≠ R H S.
So, -3 is not a root of the quadratic equation 3 x ² + 7 x + 2 = 0.

b] x ² - 6 x + 7 = 0, 2, -2, and (3 + √2).
Solution:

1) Let us take, L H S = x ² - 6 x + 7

2) We need to check whether 2, -2, and (3 + √2) are the roots of x ² - 6 x + 7 = 0 or not,

3) so put x = 2 in L H S, we get,

L H S = x ² - 6 x + 7
L H S = (2) ² - 6 (2) + 7
L H S = 4 - 12 + 7
L H S = - 8 + 7
L H S = - 1
L H S = - 1 ≠ 0
L H S ≠ R H S.
So, 2 is not a root of the quadratic equation x ² - 6 x + 7 = 0

4) so put x = - 2 in L H S, we get,

L H S = x ² - 6 x + 7
L H S = (- 2) ² - 6 (- 2) + 7
L H S = 4 + 12 + 7
L H S = 16 + 7
L H S = 23
L H S = 23 ≠ 0
L H S ≠ R H S.
So, - 2 is not a root of the quadratic equation x ² - 6 x + 7 = 0

5) so put x = (3 + √2) in L H S, we get,

L H S = x ² - 6 x + 7
L H S = (3 + √2) ² - 6 (3 + √2) + 7
L H S = (9 + 6√2 + 2)  - 18 - 6√2 + 7
L H S = 11 + 6√2   - 18 - 6√2 + 7
L H S = 18 + 6√2   - 18 - 6√2
L H S = 0
L H S =  R H S.
So, (3 + √2) is a root of the quadratic equation x ² - 6 x + 7 = 0

The next part of this topic will be published in the next Blog.