**Q1.Find the roots of the Following quadratic equations, if they exist, by the method of quadratic formula.**

Quadratic Formula/Shreedharacharya Sutra:-

**X =[-b ± √(b² – 4ac)]/2a**

*Nature of Roots:*

*Discriminant**(D)= b² – 4ac*

1. b²- 4ac < 0

Nature of Roots : Imaginary/No real root

2. b² – 4ac = 0

Nature of Roots: Real and two equal Roots

3. b² – 4ac >0

Nature of Roots: Real and two different roots

**(i) 2x² – 3x + 5 = 0**

*Discriminant(D):*

b² – 4ac = (-3)² – 4×2×5

⇒ b² – 4ac = 9 – 40

⇒ b² – 4ac = -31 < 0

Nature of roots: No Real and No Solution

**(iii) 2x² – 6x + 3 = 0**

*Discriminant(D):*

b²- 4ac = (-6)² – 4×2×3

⇒ b²- 4ac = 36 – 24

⇒ b²- 4ac = 12>0

Nature of roots: Two Real and Different Roots

By Quadratic Formula:

x = [-b ± √(b² – 4ac)]/2a

⇒ x = [-(-6) ± √12]/2×2

⇒ x = [ 6 ± 2√3]/4

⇒ x = 2[ 3 ± √3]/4

x = (3+√3)/2 ; x = (3-√3)/2

x = (3+√3)/2, (3-√3)/2

**(iii) **

**Solution:**

**Discriminant:**

b² – 4ac = (-4√3)² – 4×3×4

⇒ b² – 4ac = 48 – 48

⇒ b² – 4ac = 0

Nature of Roots: Two Real and Equal Roots

By Quadratic Formula:

x= [-b ± √(b² – 4ac)]/2a

⇒ x = [-(-4√3) ± √0]/2×3

⇒ x = [-(-4√3) ± √0]/6

⇒ x = 4√3 ± 0]/6

x = 4√3 – 0/6 ; x = 4√3 + 0/6

⇒ x = 4√3/6 ; x = 4√3/6

⇒ x = 2√3/3 ; x = 2√3/3

⇒ x = 2/√3 ; x = 2/√3

x = 2/√3 ,2/√3

**Q.2 Find the value of k for each of the following by Quadratic Equations, so that they have two equal roots.**

**(i) 2x²+ k x + 3 = 0**

**Solution:**

Condition For two real roots.

b² – 4ac = 0

⇒ k² – 4×2×3 = 0

⇒ k² – 24 = 0

⇒ k² = 24

⇒ k = ±√24

⇒ k = ±2√6

(ii) k x(x -2) + 6 = 0

**Solution:**

k x(x -2) + 6 = 0

⇒ k x² – 2kx + 6 = 0

Condition for two real and equal roots

b² – 4ac = 0

⇒ (2k)² – 4×k×6 = 0

⇒ 4k² – 24k = 0

⇒ 4k(k – 6) = 0

4k = 0 ; k – 6 = 0

⇒ k = 0/4 ; k = 6

⇒ k = 0 , 6

**Q.3 Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.**

**Solution:**

Let the Breadth of mango grove = x m and

Length = Area of Rectangle

⇒ 2 x × x = 800

⇒ 2 x² = 800

⇒ x² = 800/2

⇒ x² = 400

⇒ x = √400

⇒ x = 20

Hence length is 2x = 2×20 = 40 m

Breadth = 20 m

**Q.4 Is the following situation possible? If so, determine their present ages. ****The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48. **

Let the present age of 1st friend be = x years

Age of 2nd friend be = (20 – x)years

Before 4 years

1st Friend age was = (x – 4) years

2nd Friend age was = 20 – x – 4= (16 – x) years

According to condition:

Product of their Ages = 48

(x – 4)(16 – x) = 48

⇒ x(16 – x) – 4(16 – x) = 48

⇒ 16x – x² – 64 + 4x = 48

⇒ 20x – x² – 64 – 48 = 0

⇒ – x² + 20x – 112 = 0

⇒ – ( x² – 20x + 112) = 0

⇒ x² – 20x + 112 = 0

Discriminant:

D = b² – 4ac

⇒ D = (-20)² – 4×1×112

⇒ D = 400 – 448

⇒ D = -48 < 0

Nature of roots ; No real roots.

No this situation is not possible.

**Q.5 Is it possible to design a rectangular park of perimeter 80 m and area 400 m²? If so, find its length and breadth.**

**Solution: **

2( Length + Breadth ) = Perimeter of Rectangular park

⇒ 2 (L + B) = 80

⇒ L + B = 80/2

⇒ L + B = 40 ———(i)

Let Length be = x meters

Length + Breadth = 40

⇒ x = 40 – x

Length × Breadth = Area of Rectangular Park

⇒ x × (40 – x) = 400

⇒ 40x – x² = 400

⇒ x² – 40x + 400 = 0

⇒ x² – (20 + 20)x + 400 = 0

⇒ x² – 20x – 20x + 400 = 0

⇒ x (x – 20) – 20(x – 20) = 0

⇒ (x – 20)(x – 20) = 0

x = 20 ; x = 20

Hence Length of Park = 20 m

Breadth of Park = 20 m