What is a Quadratic EquationAn equation of the form ax ^{2} + bx + c = 0 where a≠0 is called a quadratic equation. Features of a quadratic equation
 It consists of one variable.
The variable in the above case is x.
 The variable is raised to different powers.
The powers to which it is raised is 0, 1 and 2. The result is x^{0} or 1, x^{1} or x and x^{2}.
 The variables raised to different powers are multiplied by some constants.a,b and c.
c×1,bx and ax^{2}.
 Each term obtained is added.
ax^{2} + bx + c.
 The highest degree of the variable is 2 with coefficient not equal to zero i.e a≠0.
 The whole expression is equated to zero.
ax^{2} + bx + c = 0
Root of a quadratic EquationA value which satisfies the quadratic equation is called the root of the quadratic equation. If α is a root of ax ^{2} + bx + c = 0 then aα ^{2} + bα + c = 0. There are two roots of a quadratic equation. Solving a quadratic equation by factorization
 Reduce the equation to the quadratic form i.e. in the form ax^{2} + bx + c = 0.
 Factorize the expression on the left hand side.
 Put each factor to zero and solve.
Solving a quadratic equation by formulaThe roots of a quadratic equation can be obtained by using the formula
SolutionsThe equation is ax ^{2} + bx + c = 0. It can be written as a(x ^{2} + (b/a)x + (c/a)) = 0 As a≠0 because if it is zero then the equation is not quadratic. So, (x ^{2} + (b/a)x + (c/a)) = 0
Solution 1A quadratic equation can be solved by removing the second term in ax ^{2} + bx + c = 0. This can be achieved by completing the square. (x ^{2} + (b/a)x + (c/a)) = 0 Completing the square (x ^{2} + (2b/2a)x + b ^{2}/4a ^{2}  b ^{2}/4a ^{2} + (c/a)) = 0 (x ^{2} + (2b/2a)x + b ^{2}/4a ^{2}) = b ^{2}/4a ^{2}  (c/a) (x + (b/2a)) ^{2} = b ^{2}/4a ^{2}  (c/a) (x + (b/2a)) ^{2} = (b ^{2}  4ac)/4a ^{2} (x + (b/2a)) = ±√[(b ^{2}  4ac)/4a ^{2}] x = (b/2a) ±√[(b ^{2}  4ac)/4a ^{2}] x = [b ± √(b ^{2}  4ac)]/2a
The two roots are x = [b + √(b ^{2}  4ac)]/2a x = [b − √(b ^{2}  4ac)]/2a
Solution 2(x ^{2} + (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 = ± √[(b ^{2}  4ac)/a ^{2}] 2x = (b/a) ± √[(b ^{2}  4ac)/a ^{2}] x = [b ± √(b ^{2}  4ac)]/2a
The two roots are x = [b + √(b ^{2}  4ac)]/2a x = [b − √(b ^{2}  4ac)]/2a
Solution 3(x ^{2} + (b/a)x + (c/a)) = 0 As (x  A)(x  B) = x ^{2}  (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} = (b ^{2}  4ac)/a ^{2}(B  A) = + √[(b ^{2}  4ac)/a ^{2}] B + A + B  A = 2B = (b/a) + √[(b ^{2}  4ac)/a ^{2}] B = (b/2a) + √(b ^{2}  4ac)/2a B = [b + √(b ^{2}  4ac)]/2a A = B + A  B = (b/2a)  √[(b ^{2}  4ac)]/2a A = [b  √(b ^{2}  4ac)]/2a
The two roots are x = [b + √(b ^{2}  4ac)]/2a x = [b − √(b ^{2}  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 x ^{2} (A+B)x + AB = 0.
For example, 4 and 5 are roots of the equation x^{2} (4+5)x + 4×5 = 0 x^{2}  9x + 20 = 0
