Quadratic Equations - Sibanand Pattnaik
QA/DILR Mentor | Be Legend
Any equation of degree 2 is known as a quadratic equation.
General form is ax^2+bx + c = 0
The numbers a, b are called the coefficients of this equation and c is the constant.
The possible values of x which satisfy the quadratic equation are called the roots of the quadratic equation. Quadratic equation will have two roots either real or imaginary. We normally denote them as α and β.
A root of the quadratic equation is a number such that pα ^2 + qα + r = 0 or pβ^ 2 + qβ + r = 0.
If α & β are roots of the quadratic equation ax^2+bx + c = 0, then (x - α) and (x -β) are factors of ax^2+bx + c = 0
OR ax^2+bx + c = (x - α)(x -β)
Properties of roots
f(x) = ax^2 + bx + c = 0
The sum of the roots= α + β = -b/a.
The product of roots = α β = c/a
If α and β are the roots of the equation ax^2 + bx + c = 0, then we can write
ax^2 + bx + c = x^2 - (α + β) x + αβ = x^2 + (sum of the roots) x + product of the roots = 0
Or, ax^2 + bx + c = a (x – α) (x – β) = 0
Given is the quadratic equation ax^2 + bx + c = 0, where a ≠ 0.
(b^2 – 4ac) is also known as Discriminant (D)
If D = 0, then √(b^2-4ac)= 0. So, the roots will be real and equal.
If D > 0, then √(b^2-4ac)> 0. So, the roots will be real and distinct.
If D < 0, then √(b^2-4ac) is not real. So, the roots will not be real.
If D is a perfect square (including D = 0) and a, b and c are rational, then the roots will also be rational.
Descartes' Rule of Signs of Roots
The maximum number of positive roots of any equation is equal to the change of signs from positive (+ve) to negative (-ve) and from negative (-ve) to positive (+ve).
Solving by factorization
Ax^2 + Bx + C =0
We have to write B as the sum of 2 numbers say P and Q such that the product of P and Q is equal to the product of A and C
B = (P + Q)
A * C = P * Q
5x^2 -2x -4 = 0
So P * Q = 5 * (-4) = -20
And P + Q = -2
By trial and error we get - 10 and 2
5x^2 -10x + 2x -4 = 0
5x(x-2) + 2(x-2) =0
Solving by Formula
Assuming that α and β are the roots of the equation ax^2 + bx + c = 0, where a ≠ 0.
Then α = (-b+√(b^2-4ac))/2a and β = (-b-√(b^2-4ac))/2a
Minimum Value of Quadratic Equations
Ax^2 + Bx + C = 0
OR (sqrtA * x)^2 + (2 * (sqrtA * x) * B/(2 * (sqrtA)) ) + (B/(2 * (sqrtA * x)^2 + C = 0
OR (sqrtA * x + (B/(2 * (sqrtA)) ^2 + (C - (B/(2 * (sqrtA * x)^2 ) = 0
Now (sqrtA * x + (B/(2 * (sqrtA*x)) ^2 will always be positive as it’s a square term so its minimum value will be 0
SO the minimum value of the expression (sqrtA * x + (B/(2 * (sqrtA * x)) ^2 + (C - (B/(2 * (sqrtA * x)^2 ) is equal to:
(C - (B/(2 * (sqrtA * x)^2 )
And this value is achieved when (sqrtA * x + (B/(2 * (sqrtA)) is 0
Or x = -B/2A
So the minimum value of a quadratic equation is (C - (B/(2 * (sqrtA * x)^2 ))
And this value is achieved when x = -B/2A
When A (coefficient of x^2) is positive
We cant find the minimum value but not the maximum value
The above post also helps us relate quadratic equation to graphs:
We all know the graph of a quadratic equation is a parabola
Graph of (x+a)^2 will be same, only shift a place to the left on x axis
Graph of (x-a)^2 will be same, only shift a place to the right on x axis
Graph of (x)^2 + b will be same, only shift b place up on y axis
Graph of (x)^2 - b will be same, only shift b place down on y axis
We can therefore draw the graph of the quadratic equation say (x^2 - 4x - 12) = (x-2)^2 – 16
So the graph will be graph of x^2 shifted 2 places to the right on x axis and 16 places down on the y axis
Its lowest point will be when y = -16 and x = 2
Also in the graph of (x^2 - 4x - 12), when x = 0, y = -12 so the graph will cut the y axis when y =-12
Also (x^2 - 4x - 12) = (x-6)(x+2)
So the graph will cut x axis at 6 and -2
So one positive and one negative root which is also given by the sign changes
So the graph will go down from 2nd quadrant, cut x axis at -2 and enter the 3rd quadrant , then cut y axis at -12 and enter the 4th quadrant, then reach its lowest point when y =-16 and x=2 and then go up and cut x axis at 6 and into the 1st quadrant
So if the value of c is positive therefore the lowest point of the graph is positive so the graph never reaches x axis so there are no real roots (all this is when the graph is U shaped i.e. a is positive)
When A (coefficient of x^2) is negative
We cant find the maximum value but not the minimum value
Everything will be same as above except in the inverted form
So if C is negative so the equation will not have any real roots
Also check the below videos for some good concepts