Class10 NCRT Quadratic Equations Exercise – 4.4 pdf || UP Board

 Quadratic Equations
Exercise – 4.4

 

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them

 

(i) 2x² – 3x + 5 = 0

The given equation is of the type ax² + bx + c = 0, where a = 2, b = -3 and c = 5

Therefore, b² – 4ac = (-3)² – 4 × 2 × 5 = 9 – 40 = - 31 < 0

Hence no real root exists.

 

(ii) 3x² - 4√3 x + 4 = 0

The given equation is of the type ax² + bx + c = 0, where a = 3, b = -4√3 and c = 4

Therefore, b² – 4ac = (-4√3)² – 4 × 3 × 4 = 48 – 48 = 0

Hence, the given equation has real and equal roots.

Or x = (2√3)/3, x = (2√3)/3

So the roots of the given quadratic equation are (2√3)/3 and (2√3)/3.

 

(iii) 2x² – 6x + 3 = 0

The given equation is of the type ax² + bx + c = 0, where a = 2, b = -6 and c =3

Therefore, b² – 4ac = (-6)² – 4 × 2 × 3 = 36 – 24 = 12 > 0

Hence, the given equation has real and different roots.

Or x = (3 + √3)/2, x = (3 - √3)/2

So the roots of the given quadratic equation are (3 + √3)/2 and (3 - √3)/2.

 

2.Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) 2x² + kx + 3 = 0

The given equation is of the type ax² + bx + c = 0, where a = 2, b = k and c = 3

Therefore, b² – 4ac = (k)² – 4 × 2 × 3 = k² – 24

Since both the roots of the equation are equal, so k² – 24 = 0

k² = 24

k = ±√24

k = ±2√6

 

(ii) kx(x – 2) + 6 = 0

kx² – 2kx + 6 = 0

The given equation is of the type ax² + bx + c = 0, where a = k, b = -2k and c = 6

Therefore, b² – 4ac = (-2k)² – 4 × k × 6 = 4k² – 24k

Since both the roots of the equation are equal, so 4k² – 24k = 0

4k(k – 6) = 0

4k = 0, (k – 6) = 0

k = 0, k = 6

But k ≠ 0 because it does not satisfy the equation kx(x – 2) + 6 = 0.

So, k = 6

 

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.

 

Let the breadth of the mango grove = x m

Hence, length of the mango grove = 2x m

and area = x × 2x = 2x²

According to Question,

2x² = 800

x² = 400

x = ±20

Since the breadth of the mango grove cannot be negative, so the breadth of the mango grove = 20 m

and length of the mango grove = 2 × 20 = 40 m

 

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 age of the first friend = x years

Therefore, age of second friend = 20 – x years

Four years ago the age of the first friend = x – 4 years

Therefore, age of the second friend four years ago = 20 – x – 4 = 16 – x years

According to Question,

(x – 4)(16 – x) = 48

16x – x² – 64 + 4x = 48

x² – 20x + 112 = 0

The equation is of the type ax² + bx + c = 0 , where a = 1, b = -20 and c = 112

Therefore, b² – 4ac = (-20)² – 4 × 1 × 112 = 400 – 448 = -48 < 0

Hence no real root exists.

Therefore, this situation is not possible.

 

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

 

Let the length of the park b = x

Perimeter = 80 m

Therefore, breadth of the park = 40 – x         

[Because Perimeter = 2(Length + breadth)

80 = 2(x + breadth)

x + breadth = 40

breadth = 40 – x]

According to Question,

Area =  x(40 – x) = 400

40x – x² = 400

x² – 40x + 400 = 0

x² – 20x – 20x + 400 = 0

x (x – 20) -20(x – 20) = 0

(x – 20)(x – 20) = 0

Or (x – 20)² = 0

(x – 20) = 0

x = 20

Hence, length of the park =  20 m

Therefore, breadth of the park =  40 – x = 40 – 20 = 20 m

 

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