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Revision Notes on Quadratic Equations
In order to solve a quadratic equation of the form ax2 + bx + c, we first need to calculate the discriminant with the help of the formula D = b2 – 4ac.
The solution of the quadratic equation ax2 + bx + c= 0 is given by x = [-b ± √ b2 – 4ac] / 2a
If α and β are the roots of the quadratic equation ax2 + bx + c = 0, then we have the following results for the sum and product of roots:
α + β = -b/a
α.β = c/a
α – β = √D/a
It is not possible for a quadratic equation to have three different roots and if in any case it happens, then the equation becomes an identity.
Nature of Roots:
Consider an equation ax2 + bx + c = 0, where a, b and c ∈ R and a ≠ 0, then we have the following cases:
D > 0 iff the roots are real and distinct i.e. the roots are unequal
D = 0 iff the roots are real and coincident i.e. equal
D < 0 iffthe roots are imaginary
The imaginary roots always occur in pairs i.e. if a+ib is one root of a quadratic equation, then the other root must be the conjugate i.e. a-ib, where a, b ∈ R and i = √-1.
Consider an equation ax2 + bx + c = 0, where a, b and c ∈Q and a ≠ 0, then
If D > 0 and is also a perfect square then the roots are rational and unequal.
If α = p + √q is a root of the equation, where ‘p’ is rational and √q is a surd, then the other root must be the conjugate of it i.e. β = p - √q and vice versa.
x2 – (Sum of roots)x + (Product of roots) = 0.
So if α and β are the roots of equation then the quadratic equation is
x2 – (α + β)x + α β = 0
For the quadratic expressiony = ax2 + bx + c, where a, b, c ∈ R and a ≠ 0, then the graph between x and y is always a parabola.
If a > 0, then the shape of the parabola is concave upwards
If a < 0, then the shape of the parabola is concave upwards
Inequalities of the form P(x)/ Q(x) > 0 can be easily solved by the method of intervals of number line rule.
The maximum and minimum values of the expression y = ax2 + bx + c occur at the point x = -b/2a depending on whether a > 0 or a< 0.
y ∈[(4ac-b2) / 4a, ∞] if a > 0
If a < 0, then y ∈ [-∞, (4ac-b2) / 4a]
The quadratic function of the form f(x, y) = ax2+by2 + 2hxy + 2gx + 2fy + c = 0 can be resolved into two linear factors provided it satisfies the following condition: abc + 2fgh –af2 – bg2 – ch2 = 0
In general, if α1,α2, α3, …… ,αn are the roots of the equation
f(x) = a0xn +a1xn-1 + a2xn-2 + ……. + an-1x + an, then
1.Σα1 = - a1/a0
2.Σ α1α2 = a2/a0
3.Σ α1α2α3 = - a3/a0
Σ α1α2α3 ……αn= (-1)n an/a0
Every equation of nth degree has exactly n roots (n ≥1) and if it has more than n roots then the equation becomes an identity.
If there are two real numbers ‘a’ and ‘b’ such that f(a) and f(b) are of opposite signs, then f(x) = 0 must have at least one real root between ‘a’ and ‘b’.
Every equation f(x) = 0 of odd degree has at least one real root of a sign opposite to that of its last term.
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Solved Examples on Quadratic Equations...