Quadratic Equations Introduction

Definition
  1. A quadratic is a polynomial whose highest degree is 2.
  2. aX2 + bX + c = 0 where a, b, c are numbers and a ≠ 0. [Standard Form of Quadratic equation]
  3. The co-efficient of x2 is called the leading co-efficient.
  4. Root : A solution of a quadratic equation is known as Root. There is always two roots of a quadratic equations it may be equal, different or complex.
Quadratic Equation

Let aX2 + bX + c = 0 where a, b and c are real numbers and a ≠ 0.

$\text{x} = \cfrac{\text{-b} \pm \sqrt{\text{b}^2 - 4\text{ac}}}{2\text{a}}$
Number of roots

Discriminant : In Qudratic formula (b2 - 4ac) is called the Discriminant, where a, b and c are real numbers, a ≠ 0. It is represented by D.

  1. When D = b2 - 4ac is positive, then the roots are real and unequal.
  2. When D = b2 - 4ac is zero, then roots are real and equal.
  3. When D = b2 - 4ac is negative, then roots are unequal and imaginary.
  4. When D = b2 - 4ac is positive and perfect square, then the roots are real,rational and unequal.
  5. When D = b2 - 4ac is positive but not perfect square, then roots are real, irrational and unequal.
  6. When D = b2 - 4ac is perfect square and a or b is irrational, then roots are irrational.

1. Find the number of roots of x2 + 3x - 10 = 0

Discriminant, D = b2 - 4ac

$=(-3)^2-4 \times 1 \times (-10)$
$=9+40=49 >0 $

Here, D > 0 and perfect square. Hence the equation will have real, rational and unequal roots.

Let's find the root by solving the quadratic equation.

$ \because \text{x}^2+3\text{x}-10=0 \\ \Rightarrow \text{x}^2+5\text{x}-2\text{x}+2=0 \\ \Rightarrow \text{x}(\text{x}+5)-2(\text{x}+5)=0 \\ \Rightarrow (\text{x}+5)(\text{x}-2)=0 \\ \Rightarrow \text{x}=2 \ or \ -5 $

The solution is $\text{x}=2,-5$

2. Find the number of roots of x2 - 6x + 9 = 0

Discriminant, D = b2 - 4ac

$=(-6)^2-4 \times 1 \times 9$
$ =36-36=0 $

Here, D = 0. Hence the equation will have one real and equal roots.

Let's find the root by solving the quadratic equation.

$ \because \text{x}^2-6x+9=0 \\ \Rightarrow \text{x}^2-3\text{x}-3\text{x}+9=0 \\ \Rightarrow \text{x}(\text{x}-3)-3(\text{x}-3)=0 \\ \Rightarrow (\text{x}-3)^2=0 \\ \Rightarrow \text{x}=3 $

The solution is $\text{x}=3$

3. Find the number of roots of x2 + 3 =0

Discriminant, D = b2 - 4ac

$=(0)^2-4 \times 1 \times 3 \\= 0-12=-12 $

Here, D < 0. Hence the equation will have unequal and imaginary.

Let's find the root by solving the quadratic equation.

$\because \text{x}^2+3=0 \\ \Rightarrow \text{x} = \cfrac{\text{-b} \pm \sqrt{\text{b}^2 - 4\text{ac}}}{2\text{a}} \\ \Rightarrow \text{x} = \cfrac{0 \pm \sqrt{0^2 - 4\times 1 \times 3}}{2\times 1} \\ \Rightarrow \text{x} = \cfrac{\sqrt{-12}}{2\times 1} \\ \Rightarrow \text{x} = \cfrac{\pm 2i\sqrt{3}}{2} \\ \Rightarrow \text{x} = \pm i\sqrt{3}$
Sum and Products of the roots

Let p and q are the roots of a quadratic equation ax2 + bx + c = 0 where a ≠ 0

Sum of the roots = p + q = − b/a

Product of the roots = pq = c/a

Example : Find the sum and product of the roots for the given equation x2 + 3x - 10 = 0

Let p and q are the roots of the given equation, then Sum of the roots = p + q = − b/a = -3/1 = -3 and Product of the roots = pq = c/a = -10/1 = -10