Definition:
An equation whose degree is ‘2’ is called as quadratic equation.
It can be written in the form of
ax2 + bx + c = 0
Here, a, b and c are real numbers
a ≠ 0
Properties:
1. A quadratic equation has two roots. (root can be called as the solution of quadratic equation)
2. These two roots individually when put in the quadratic equation it satisfies the given quadratic equation.
For e.g. for quadratic equation 2x2 – 3x – 9, x = 3 satisfies the given equation so therefore x = 3 is one of the root solution of the quadratic equation.
From the given equation nature of the roots can be found out such as real, imaginary, distinct etc.
A. Methods of finding roots of quadratic equation Solution of quadratic equation
This method is based on the concept that if product of two numbers then at least one of the number is zero.
For quadratic equation,
ax2 + bx + c = 0
we will expand the middle term ‘bx’ in two terms and thereafter taking common multiple out of it. The equation will be reduced to following step.
(px + q)(rx + s) = 0
Here, either (px + q) = 0 or (rx + s) = 0
So two roots (two value of x) can be found.
Let’s understand it by example.
Case–1
when a = 1
x2 + 11x + 21 = 0
we will find two multiplies of 21 in such a way that its addition or subtraction is 11.
x2 + 11x + 21 = 0
x2 + 7x + 3x + 21 = 0
x(x + 7) + 3(x + 7) = 0
(x + 7) (x + 3) = 0
x + 7 = 0 OR x + 3 = 0
x = – 7 OR x = – 3
Solution of quadratic equation,
ax2 + bx+ c = 0 can be find using the following formula,
x=(-b±√(b2-4ac))/2a
i.e. x=(-b+√(b2-4ac))/2a OR x=(-b-√(b2-4ac))/2a
Here, the term ‘b^2-4ac’ is called as ‘Discriminant’ and it defines the nature of root.
For e.g. x2 - 15x + 54 = 0
By comparing it with,
ax2 + bx + c = 0
a = 1, b = - 15, c = 54
Let discriminant D = b2-4ac
= (-15)2 – 4(1)(54)
= 225 – 216
D = 9
x =(-b±√(b2-4ac))/2a
=(-15±√9)/(2(1))
= (-15+√9)/2 OR = (-15-√9)/(2(1))
=(-6)/2 OR = (-18)/2
x = -3 OR x = -9
First method of solving quadratic equation will be factorization method but sometimes these are some equations in which we can not find the factors of constant term ‘c’ such that its sum is middle co-efficient ‘b’ such as,
x2 + 9x + 2 = 0
x2 + 21x + 11 = 0 etc.
To solve such problems we will use completing square method.
In these method,
We will add a suitable term to (ax2 + bx) to make a completing square
For example x2 + 8x + 2 = 0
In these equation if we add ‘16’ on the both the side of equation.
x2 + 8x + 2 + 16 = 0 + 16
x2 + 8x + 16 + 2 = 16
(x + 4)2 + 2 – 16 = 0
(x + 4)2 – 14 = 0
(x + 4)2 – (√14 )2 = 0
(x + 4 + √14 ) (x + 4 – √14) = 0
(x + 4 + √14) = 0 OR (x + 4 – √14) = 0
x = – 4 – √14 OR x = – 4 + √14
How to find suitable term to make (ax2 + bx) complete square.
Let ‘k’ be the term to add (ax2 + bx) in order to make it complete square, then,
k =( b/2a )2
For e.g. x2 + 6x + 7 = 0
Here a = 1, b = 6, c = 7
k =( b/2a )2 =( 6/(2×1) )2 = (3)2 = 9
After adding and subtracting ‘9’ it will become
x2 + 6x + 9 + 7 - 9 = 0
(x + 3)2 + 7 – 9 = 0
(x + 3)2 – 2 = 0
(x + 3)2 – (√2)2 = 0
(x + 3 + √2) × (x + 3 – √2) = 0
x = – 3 – √2 OR x = – 3 + √2
Nature of roots for quadratic equations,
ax2 + bx + c = 0
Depends upon the value of a, b and c
Discriminant = d = b2 – 4ac
Case – 1 if d > 0 then the two roots will be real and distinct
Case – 2 if d = 0 then the two roots will be real and equal
Case – 3 if d < 0 then the two roots will be imaginary and distinct
For example what will be the nature of roots for quadratic equation,
x2 + x – 6 = 0
Solution:
Comparing x2 + x – 6 = 0 with ax2 + bx + c = 0
a = 1, b = 1 and c = – 6
Discriminant = d = b2 – 4ac
= 1 – 4(1)(–6)
= 1 + 24
= 25
As here d > 0,
The given quadratic equation will have two real distinct roots.
If ∝ and β are the roots of quadratic equation.
ax2 + bx + c = 0
then,
1. Sum of the roots = ∝ + β = (-b)/a
2. Product of the roots = ∝.β = c/a
D. Formation of Quadratic Equation:
Here we will form the quadratic equation when the roots (∝ and β) of a quadratic equation will be,
If ∝ and β are the two roots of a quadratic equation then,
(∝- β) (∝ + β) = 0 -> when two roots are given
x2 - (∝ + β)x + ∝.β = 0 -> when relation between the roots are
For e.g. roots of quadratic are 2 and -3. The given equation will be?
Solution:
As we know when roots of quadratic equation are given the equation will be for ∝ = 2 and β = -3
(x - ∝) (x – β) = 0
(x - 2) (x + 3) = 0
x2 + 3x – 2x – 6 = 0
x2 + x – 6 = 0
This is the required quadratic equation.
Q.1. Solve the following equation,
x + 2/3x = 7/3
Solution:
Here the above equation is not quadratic equation but if we multiply the above equation by ‘x’ it becomes the quadratic equation,
x + 2/3x = 7/3
x (x + 2/3x) = 7x/3 …multiplying both side by ‘x’
x2 + 2/3 = 7x/3
x2 – 7x/3 + 2/3 = 0
3x2 – 7x + 2 = 0 …multiplying both side by ‘3’
3x2 – 6x – x + 2 = 0
3x(x – 2) – 1(x – 2) = 0
(3x – 1) (x – 2) = 0
3x – 1 = 0 OR x – 2 = 0
x = 1/3 OR x = 2
Q.2. Solve the following equation
Solution:
x4 – 13x2 + 36 = 0
Here we can write, x4 = (x2)2
Let assumes, x2 = m, so therefore equation becomes
m2 – 13m + 36 = 0
m2 – 9m – 4m + 36 = 0
m(m – 9) – 4( m – 9) = 0
(m – 9) (m – 4) = 0
m = 9 OR m = 4
Now after resubstituting,
m = x2
x2 = 9 OR x2 = 4
x = ± 3 OR x = ± 2
Q.3. (x+3) (x+2) (x+4) (x+5) + 1 = 0
Solution:
The above equation is in the form of
(x+a) (x+b) (x+c) (x+d) + k = 0
To solve the above equation we have to make a pair of two terms
The pair will be selected on the following basis,
a + b = c + d or
a + c = b + d or
a + d = b + c
In our case,
(x+3) (x+2) (x+4) (x+5) + 1 = 0
{3 + 4 = 2 + 5}
(x+3) and (x+4) will make one pair and (x+5) and (x+2) will make another pair.
(x+3)(x+4)(x+2)(x+5) + 1 = 0
(x2 + 4x + 3x + 12) (x2 + 5x + 2x + 10) + 1 = 0
(x2 + 7x + 12) (x2 + 7x + 10) + 1 = 0
Let’s assume x2 + 7x = t
(t+12) (t+10) + 1 = 0
t2 + 10t + 12t + 120 + 1 = 0
t2 + 22t + 121 = 0
t2 + 11t + 11t + 121 = 0
t(t+11) + 11(t+11) = 0
(t+11) (t+11) = 0
t + 11 = 0 OR t + 11 = 0
t = - 11
After Resubstituting,
x2 + 7x = - 11
x2 + 7x – 11 = 0
By formula method,
a = 1, b = 7 and c = 11
b2 – 4ac = (7)2 – 4(11)(1)
= 49 – 44
= 5
x=(-b±√(b2-4ac))/2a
= (-7±√5)/2
x = (-7+√5)/2 OR x = (-7-√5)/2
Q.4. 3(x2 + 1/x2 ) – 5 (x + 1/x) + 8 = 0
Solution:
As we know,
(x+1/x)2 = x2 + 2x. 1/x + 1/x2
(x+1/x)2 = x2 + 2 + 2/x2
(x+1/x)2 – 2 = x2 + 2/x2
Put (x+1/x) = m
3(m2 – 2) – 5m + 8 = 0
3m2 – 6 – 5m + 8 = 0
3m2 – 5m + 2 = 0
3m2 – 3m – 2m +2 = 0
3m(m – 1) – 2(m – 1) = 0
m – 1 = 0 OR 3m – 2 = 0
m = 1 OR 3m = 2
∴ m = 2/3
After resubstituting
(x+1/x) = m
x+1/x = 1 and x+1/x = 2/3
x2 + 1 = x x2 + 1 = 2x/3 …multiplying by ‘x’
x2 – x + 1 = 0 3x2 + 3 = 2x
x2 – x + 1 = 0 3x2 – 2x + 3 = 0
b2 – 4ac: b2 – 4ac:
b2 – 4ac=(–1)2 –4(1)(1) b2 – 4ac = (–2)2 –4(3)(3)
= 1 – 4 = 4 – 36
= –3 = – 32
As b2 – 4ac < 0 As b2 – 4ac < 0
It has imaginary It has imaginary
Roots or no real Roots or no real roots.
Roots
SOME IMPORTANT FORMULA
Quadratic inequality can be written in the following way,
ax2 + bx + c > 0
ax2+ bx + c ≥ 0
ax2 + bx + c < 0
ax2 + bx + c ≤ 0
Steps to solve quadratic inequality:
What will be the solution of quadratic inequality
x2 + 7x + 6 ≥ 0?
Solution:
x2 + 7x + 6 = 0
x2 + 6x + 1x + 6 = 0
x2 + 6x + 1x + 6 = 0
x(x+6) +1 (x+6) = 0
(x+1) (x+6) = 0
x = –1 OR x = –6
Now, we have to find interval on number line which satisfies the above conditions,
Let’s take three points
x = –7 x = –4 x = 1
For x = –7 For x = –4 For x = 1
x2 + 7x + 6 ≥ 0 x2 + 7x + 6 ≥ 0 x2 + 7x + 6 ≥ 0
(–7)2 +7(–7)+6 ≥ 0 (–4)2 +7(–4)+6 ≥ 0 (1)2 +7(1)+6 ≥ 0
49 – 49 + 6 ≥ 0 16 – 28 + 6 ≥ 0 1+7+6≥ 0
6 ≥ 0 22 – 28 ≥ 0 14 ≥ 06
-6 ≱ 0 8 ≥ 0
The solution is x > –1 and x < –6
Steps to solve Quadratic Inequality:
Posted By:
Hardik Kaneriya
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