Notes on Quadratic Equations
Quadratic Polynomial
A polynomial, whose degree is 2, is called a quadratic polynomial. It is in the form of
p(x) = ax2 + bx + c, where a ≠ 0
Quadratic Equation
When we equate the quadratic polynomial to zero then it is called a Quadratic Equation i.e. if
p(x) = 0, then it is known as Quadratic Equation.
Standard form of Quadratic Equation
where a, b, c are the real numbers and a≠0
Types of Quadratic Equations
1. Complete Quadratic Equation ax2 + bx + c = 0, where a ≠ 0, b ≠ 0, c ≠ 0
2. Pure Quadratic Equation ax2 = 0, where a ≠ 0, b = 0, c = 0
Roots of a Quadratic Equation
Let x = α where α is a real number. If α satisfies the Quadratic Equation ax2+ bx + c = 0 such that aα2 + bα + c = 0, then α is the root of the Quadratic Equation.
As quadratic polynomials have degree 2, therefore Quadratic Equations can have two roots. So the zeros of quadratic polynomial p(x) = ax2 + bx + c is same as the roots of the Quadratic Equation ax2+ bx + c = 0.
Methods to solve the Quadratic Equations
There are three methods to solve the Quadratic Equations-
1. Factorization Method
In this method, we factorize the equation into two linear factors and equate each factor to zero to find the roots of the given equation.
Step 1: Given Quadratic Equation in the form of ax2 + bx + c = 0.
Step 2: Split the middle term bx as mx + nx so that the sum of m and n is equal to b and the product of m and n is equal to ac.
Step 3: By factorization we get the two linear factors (x + p) and (x + q)
ax2 + bx + c = 0 = (x + p) (x + q) = 0
Step 4: Now we have to equate each factor to zero to find the value of x.
\[ \Rightarrow {x^2} - 2x - 15 = 0\]
\[(x + 3)(x - 5) = 0\]
\[x + 3 = 0\,\,or\,\,x - 5 = 0\]
\[x = - 3\,\,or\,\,x = 5\]
These values of x are the two roots of the given Quadratic Equation.
2. Completing the square method
In this method, we convert the equation in the square form (x + a)2 - b2 = 0 to find the roots.
Step1: Given Quadratic Equation in the standard form ax2 + bx + c = 0.
Step 2: Divide both sides by a
\[{x^2} + \dfrac{b}{a}x + \dfrac{c}{a} = 0\]
Step 3: Transfer the constant on RHS then add square of the half of the coefficient of x i.e. \({\left( {\dfrac{b}{{2a}}} \right)^2}\) on both sides
\[{x^2} + \dfrac{b}{a}x = - \dfrac{c}{a}\]
\[{x^2} + \dfrac{b}{a}x + {\left( {\dfrac{b}{{2a}}} \right)^2} = - \dfrac{c}{a} + {\left( {\dfrac{b}{{2a}}} \right)^2}\]
Step 4: Now write LHS as perfect square and simplify the RHS.
\[{\left( {x + \dfrac{b}{{2a}}} \right)^2} = \dfrac{{{b^2} - 4ac}}{{4{a^2}}}\]
Step 5: Take the square root on both the sides.
\[x + \dfrac{b}{{2a}} = \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} \]
Step 6: Now shift all the constant terms to the RHS and we can calculate the value of x as there is no variable at the RHS.
\[x = \pm \sqrt {\dfrac{{{b^2} - 4ac}}{{4{a^2}}}} - \dfrac{b}{{2a}}\]
3. Quadratic formula method
In this method, we can find the roots by using quadratic formula. The quadratic formula is
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
where a, b and c are the real numbers and b2 – 4ac is called discriminant.
To find the roots of the equation, put the value of a, b and c in the quadratic formula.
Nature of Roots
From the quadratic formula, we can see that the two roots of the Quadratic Equation are -
\[x = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}}\,\,and\,\,x = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}\]
Or \(x = \dfrac{{ - b + \sqrt D }}{{2a}}\) ; Where D = b2 – 4ac
The nature of the roots of the equation depends upon the value of D, so it is called the discriminant.
∆ = Discriminant
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