Quadratic equation:

Quadratic equation:


Hi, welcome you all,
We all have study the quadratic equation in high school and college.
Today, I am going to discussed this topic, since the quadratic equation is not only important for scientist use, missile, satellite and various physical movement, but it is also used in general life in daily use.


Consider a man want to make a playground such that in 300 square meter area, having length is two times it’s width plus one meter extra. He told about it to an Engineer. What will be the area and length of the playground?

The Standard form of quadratic equation is,

\(\begin{array}{c}a{x^2} + bx + c = 0\\{\rm{Here}},a \ne 0\end{array}\)
The above quadratic equation contains only one variable x. which is 2nd order polynomial with single variable.
Is this equation is quadratic equation:
\(a{x^2} + by + c = 0\)

This is not a quadratic equation; but, it is 2nd order polynomial in two variable i.e. x and y.

The three ways to find the solution of quadratic equation are,

1. By factorization method
2. By square completion method
3. By roots finding method


Solution of quadratic equation by factorization method:

\({x^2} - x - 6 = 0\)
Here, multiply the coefficient of \({{x^2}}\) and constant  \(\left( { - 6} \right)\)
\(1 \times  - 6 =  - 6\)
The possible factor of -6 are,
\(\left\{ {\left( {1, - 6} \right),\left( { - 1,6} \right),\left( { - 2,3} \right),\left( { - 3,2} \right)} \right\}\)
To split the middle term use \(\left( { - 3,2} \right)\) that satisfies as,

\(\begin{array}{c}{x^2} - 3x + 2x - 6 = 0\\x\left( {x - 3} \right) + 2\left( {x - 3} \right) = 0\\\left( {x - 3} \right)\left( {x + 2} \right) = 0\\x - 3 = 0\\x = 3,\\x + 2 = 0\\x =  - 2\end{array}\)

Solution of quadratic equation by square completion method:

\(\begin{array}{c}4{x^2} - 4x + 1 = 0\\{\left( {2x} \right)^2} - 2 \cdot 2x + 1 = 0\\{\left( {2x + 1} \right)^2} = 0\\{\rm{since, }}{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\\{\rm{ hence,}}{\left( {2x - 1} \right)^2} = 0\\x = \frac{1}{2},\frac{1}{2}\end{array}\)


Solution of quadratic equation by roots finding method:

\(\begin{array}{c}a{x^2} + bx + c = 0\\a \ne 0,\\x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\\\alpha  = \frac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\\\beta  = \frac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\\\alpha  + \beta  = \frac{{ - b}}{a}\\\alpha \beta  = \frac{{{{\left( { - b} \right)}^2} - {{\left( {\sqrt {{b^2} - 4ac} } \right)}^2}}}{{{{\left( {2a} \right)}^2}}}\\ = \frac{{{b^2} - {b^2} - 4ac}}{{4{a^2}}}\\ = \frac{{4ac}}{{4{a^2}}}\\ = \frac{c}{a}\end{array}\)

Properties of quadratic equation:

\(\begin{array}{c}a{x^2} + bx + c = 0\\{\rm{Here}},a \ne 0\end{array}\)
If = 0 then the roots are equal real number, i.e. same roots.
If > 0 then the roots are different real numbers
If < 0 then the roots are different imaginary numbers.

Let us consider the following example:

\(\begin{array}{c}4{x^2} - 4x + 1 = 0\\a = 4,b =  - 4,c = 1\\x = \frac{{ - \left( { - 4} \right) \pm \sqrt {{{\left( { - 4} \right)}^2} - 4 \cdot 4 \cdot 1} }}{{2 \cdot 4}}\\ = \frac{{4 \pm \sqrt {16 - 16} }}{8}\\ = \frac{4}{8}\\ = \frac{1}{2},\frac{1}{2}\\{\rm{since, }}D = \sqrt {{{\left( { - 4} \right)}^2} - 4 \cdot 4 \cdot 1} \\ = 0\end{array}\)
Thus, the roots are equal real number.

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