CBSE Class 10 Maths - Quadratic Equations - Nature of Roots (Solved Problems) #class10maths #cbse2021 #quadraticequations #eduvictors

CBSE Class 10 Maths - Quadratic Equations - Nature of Roots (Solved Problems) 

#class10maths 

We know that A polynomial p(x) = ax² + bx + c of degree 2 is called a quadratic polynomial, then p(x) = 0 is known as quadratic equation.

There are three methods to find the solution of a quadratic equation:

1. Factorisation method

2. Method of completing the square 

3. Quadratic formula method

What is a Discriminant of Quadratic Equation ax² + bc + c = 0 ?

In general, Discriminant is represented by symbol D. 

Mathematically,

D = $D = b^2 - 4ac$

The Discriminant helps us find the nature of roots.

⑴ If D > 0 i.e. $b^2 - 4ac > 0$ then the given quadratic equation (QE) has two distinct real roots. (Note: Distinct mean unique values, real mean real numbers).

Value of roots will be :

$x = \frac{-b + \sqrt{D}}{2a}$ and $x = \frac{-b - \sqrt{D}}{2a}$

⑵ If D = 0 i.e. $b^2 - 4ac > 0$ then the QE has equal real roots.

i.e. $x = \frac{-b}{2a}$ and $x = \frac{-b}{2a}$

⑶ If D < 0 then $\sqrt{D} = \sqrt{-ve }$, there is no real roots exist.

We can say roots are imaginary or complex numbers (that you'll study in Class XI).

Sum of Roots of a quadratic equation:

For real roots, α = $\frac{-b + \sqrt{D}}{2a}$ and β = $\frac{-b - \sqrt{D}}{2a}$

Now α + β = \frac{-b + \sqrt{D}}{2a} +\frac{-b - \sqrt{D}}{2a}$

 = $\frac{-b + \sqrt{D} -b - \sqrt{D}}{2a} $

 = $\frac{-2b}{2a} $

 = $\frac{-b}{a}$

Product of roots of a quadratic equation:

For real roots, α = $\frac{-b + \sqrt{D}}{2a}$ and β = $\frac{-b - \sqrt{D}}{2a}$

Now α × β = $\frac{-b + \sqrt{D}}{2a} \times \frac{-b - \sqrt{D}}{2a} $

= $\frac{(-b)^2 - (\sqrt{D})^2}{4a^2} $

= $\frac{b^2 - D}{4a^2} $

= $\frac{b^2 -b^2 + 4ac}{4a^2}$  

= $\frac{4ac}{4a^2}$  

= $\frac{c}{a}$

For a quadratic equation ax² + bx + c = 0, 

ⅰ If sum of roots is 0, then b = 0

ⅱ If roots are reciprocal of each other then a = c

ⅲ If products of roots is 0, then c = 0.

Q1: Write the discriminant of each of the following quadratic equations. Find the nature of roots also.

x² + 4x + 3 =0 

Answer: D = b² - 4ac 

In the given QE, a = 1, b = 4, c = 3 

∴ D = 4² - 4(1)(3)

  D = 16 - 12

  D = 4

∵ D > 0, roots are distinct and real.

Q2: Write the discriminant of each of the following quadratic equation:

2x² + 4x + 5 = 0

Answer: In the given QE, a = 2, b = 4 and c = 5

  D = b² - 4ac

∴ D = 4² - 4(2)(5)

  D = 16 - 40

  D = -24

∵ D < 0, roots are non-real or imaginary.

Q3: Write the discriminant of each of the following quadratic equations and also find the nature of roots.

$\sqrt{3}x^2 -2\sqrt{2} - 2\sqrt{3} = 0$

Answer: In the given quadratic equation, a = $\sqrt{3}$, b = $-2\sqrt{2}$ and c = $- 2\sqrt{3}$

  D = b² - 4ac

∴ D = $(-2\sqrt{2})^2 - 4(\sqrt{3})(- 2\sqrt{3})$

= 8 - (-24)

= 8 + 24

= 32

∵ D > 0, roots are distinct and real.

Q4: Write the discriminant of each of the following quadratic equations and also find the nature of roots.

x² + 6x + 9 = 0

Answer:

In the given QE, a = 1, b = 6 and c = 9

  D = b² - 4ac

∴ D = 6² - 4(1)(9)

  D = 36 - 36

  D = 0

∵ D = 0, roots are real and equal.

Q5: Find the value of k, so that the quadratic equation

(k + 1) x² – 2 (k — 1) x + 1 = 0 has equal roots.

Answer: In the given QE, 

a = k + 1, b = -2(k - 1) and c = 1

For equal roots Disrciminant D = 0

∴  D = b² - 4ac = 0

⇒ (-2(k - 1))² - 4(k + 1)(1) = 0

  4(k-1)² - 4k - 4 = 0

  4[(k-1)² -k - 1] = 0

  4[k² -2k + 1 -k -1] = 0 

  4[k² -3k] = 0

⇒ 4k(k-3) = 0

⇒ 4 ≠ 0, k = 0 or k - 3 = 0

⇒ k = 0 or k = 3

∴ Values of k are 0, 3

 

👉See Also:

CH 4: Quadratic Equations (Ex 4.1) 
CH 4: Quadratic Equations (NCERT Ex 4.2)
CH 4: Quadratic Equations (Summary)
CH 4: 10 Question Patterns Commonly Asked in Question Paper