Quadratic Equation


Roots of a Quadratic Equation:

The standard form of a quadratic equation is: ax2 + bx + c = 0.
The roots of this quadratic are:
x = [–b – √ (b2 – 4ac)]/2a,
x = [–b + √ (b2 – 4ac)]/2a

in the roots,
b2 – 4ac is called the Discriminant of the quadratic equation.

1. Nature of roots of a quadratic equation:

Depending on this discriminant, b2 – 4ac, the roots of a quadratic equation are:
1. Real and distinct, if b2 – 4ac > 0
2. Real and Equal, if b2 – 4ac = 0
3. Imaginary (not real), if b2 – 4ac < 0

Imp Tip:
Each of the two roots is x = - (b/2a), when b2 – 4ac = 0

(i.e. when the two roots of a quadratic equation are real and equal)
We will not consider the 3rd case in this lesson.

Examples:
Examine the nature of roots of the quadratic equations:
1. x2 - 12x + 36 = 0
Solution:

To examine the nature of roots of a quadratic is to find if the roots are:
Real or Equal or Not Real.
For this, the discriminant b2 – 4ac is used.
In the quadratic, a = 1, b = -12, c = 36.
b2 – 4ac = (-12) – 4 ×1 × 36 = 144 – 144 = 0,

Since the discriminant b2 – 4ac = 0:
The roots of the quadratic are real and equal. And each of the two roots is:
x = - (b/2a) = - (-12)/2 ×1= 12/2 = 6
So, x = 6 is the only root of the quadratic equation x2 – 12x + 36 = 0

2. x2 + x + 1 = 0
Solution: in the quadratic equation: a = 1, b = 1 and c = 1.
Let us use the discriminant: b2 – 4ac to find the nature of the roots.
b2 – 4ac = 12 - 4×1×1 = 1 – 4= - 3,
since b2 – 4ac < 0 , the roots of the quadratic are not real (i.e. imaginary or complex)

3. x2 - 5x + 6 = 0
Solution: in the quadratic equation: a = 1, b = - 5. c = 6
b2 – 4ac = (-5)2 – 4×1×6 = 25 – 24 = 1
b2 – 4ac > 0, therefore roots are real and distinct.
To find the two roots of the quadratic equation:
x2 - 5x + 6 = 0, use the quadratic formula method.
x = [-(-5) – 1]/2 × 1, x = [4]/2 = 2 or
x = [-(-5) + 1]/2 × 1, x = [6]/2 = 3

4. Find p so that the roots of the quadratic equation: 9x2 – 24x + p = 0 are real and different.
Solution:
in the quadratic: a = 9, b = -24, c = p
For roots to be real and distinct: b2 - 4ac > 0 i.e.
(-24)2 – 4 × 9 × p > 0,
576 – 36p > 0,
576 > 36p,
36p < 576, p < 16

If p < 16, roots are real and distinct
If p <= 16, then roots are real and <=
Equal, if p = 16, or distinct, if p < 16

Note: So, when it is b2 – 4ac = 0, the roots are:
Real and Equal when b2 - 4ac = 0, or
Real and distinct when b2 - 4ac > 0

5. For what value of k, the roots of the quadratic equation: kx(x – 2) + 6 = 0 are real and equal.
Solution:
Simplify the quadratic as:
kx(x – 2) + 6 = 0 = kx2 – 2kx + 6 = 0
Now, a = k, b = -2k, c = 6
Roots of the quadratic are real and equal if b2 – 4ac = 0,
(-2k)2 – 4 × k × 6 = 0, 4k2 – 24k = 0, 4k (k – 6) = 0,
Either 4k = 0 or k – 6 = 0, k = 0 or k = 6.
So, if k = 0 or 6, roots of the quadratic are real and equal.

2. Sum and Product of roots in a quadratic equation:

Call the two roots in the quadratic equation ax2 + bx + c = 0 as “p” and “q”
Then, sum of the roots: p + q = - (b/a) and
product of the roots: pq = c/a

Examples: What is the sum and product of the roots in the quadratic equations?
1. x2 – 12x + 36 = 0

Solution: in the quadratic equation, a = 1, b = -12, c = 36
sum of the roots = - (-12/1) = 12,
Product of the roots = 36/1 = 36

2. 9x2 – 12(k + x) – 36k = 0
Solution: collect x terms and constant terms at one place.
9x2 – 12k – 12x – 36k = 0,
9x2 – 12x – 12k – 36k = 0,
9x2 – 12x – 12(k + 36) = 0,

In the standard form now, a = 9, b = -12, c = -12(k +36)
Sum of the roots = -b/a = -(-12/9) = 12/9 = 4/3
Product of roots = c/a = -12(k +36)/9 = -4(k +36)/3

3. Find a so that sum and product of roots of the quadratic equation:
ax2 + 2x + 3a = 0 are equal.
Solution: in the quadratic a = a, b = 2, c = 3a
Sum of roots = -b/a = -(2/a) and
product of roots = c/a = 3a/a = 3
as sum and product of roots are equal,
-2/a = 3, a = -2/3
3. How to form equations when roots are given:
The quadratic equation whose roots are “p” and “q” is:
x2 – x(p + q) + pq = 0

Examples:

1. write the quadratic equation whose roots are 4 and 6.
Solution:
sum = 10 and product = 24
So the quadratic equation is
x2 – x (4 + 6) + 4 ×6 = 0, i.e. x2 – 10x + 24 = 0

Signs of roots of a quadratic equation:

Consider the quadratic equation ax2 + bx + c = 0

1. Both roots are positive:
If a and b are opposite in sign and a and c are same in sign.
2. Both roots are Negative:
If a, b and c are all of same sign
3. Roots are of opposite signs:
If a and c are of opposite signs.
4. Roots equal but opposite in signs.
If b = 0

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