HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Haryana State Board HBSE 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.4 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
(i) 2x² – 3x + 5 = 0
(ii) 3x² – 4√3x + 4 = 0
(iii) 2x² – 6x + 3 = 0
Solution :
(i) The given equation is :
2x² – 3x + 5 = 0
Comparing the given equation with the general
form of quadratic equation, ax² + bx + c = 0, we get
a = 2, b = – 3, c = 5
∴ D = b² – 4ac
⇒ D = (- 3)² – 4 × 2 × 5
⇒ D = 9 – 40
⇒ D = – 31.
∵ D < 0
The real roots do not exist.

(ii) The given equation is :
3x² – 4√3x + 4 = 0
Comparing the given equation with the general form of quadratic equation, ax² + bx + c = 0, we get
a = 3, b = – 4√3 , c = 4
∴ D = b² – 4ac
⇒ D = (- 4√3)² – 4 × 3 × 4
⇒ D = 48 – 48
⇒ D = 0
The roots of the equation are real and equal.
x = \(-\frac{b}{2 a} \text { and }-\frac{b}{2 a}\)
x = \(\frac{4 \sqrt{3}}{2 \times 3} \text { and } \frac{4 \sqrt{3}}{2 \times 3}\)
x = \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)
Hence, the roots of the equation are \(\frac{2}{\sqrt{3}} \text { and } \frac{2}{\sqrt{3}}\)

(iii) The given equation is :
2x² – 6x + 3 = 0
Comparing the given equation with the general form of quadratic equation ax2 + bx + c = 0, we get a = 2, b = – 6, c = ∴ D = b² – 4ac
⇒ D = (- 6)² – 4 × 2 × 3
⇒ D = 36 – 24
⇒ D = 12
∵ D > 0
The roots of the equation are real and distinct.

Hence, the roots of the equation are \(\frac{3+\sqrt{3}}{2}\) and \(\frac{3-\sqrt{3}}{2}\).

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots :
(i) 2×2 + kx + 3 = 0
(ii) kx(x – 2) + 6 = 0.
Solution:
(i) The given equation is :
2x² + kx + 3 = 0
Comparing the given equation with general form of the quadratic equation ax2 + bx + c = 0, we get
a = 2, b = k, c = 3
D = b² – 4ac
⇒ D = k² – 4 × 2 × 3
⇒ D = k² – 24
Since, the roots of the given equation are equal.
Therefore, D = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = ± √24
⇒ k = ± 2√6
Hence, k = ± 2√6

(ii) The given equation is : kx (x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Comparing the given equation with the general form of quadratic equation ax² + bx + c = 0, we get
a = k, b = – 2k, c = 6
D = b² – 4ac
⇒ D = (- 2k)² – 4 × k × 6
⇒ D= 4k² – 24k
Since the roots of the given equation are equal.
Therefore, D = 0
⇒ 4k² – 24k = 0
⇒ 4k(k – 6) = 0
⇒ 4k = 0 or & k – 6 = 0
⇒ k = 0 or k = 6
But k = 0 [Not possible]
Hence, k = 6.

Question 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 the rectangular grove be x m.
Then length of rectangular grove = 2x m
According to question,
2x × x = 800
[∵ Area of rectangle = Length × Breadth]
⇒ 2x² = 800
⇒ x² = \(\frac{800}{2}\)
⇒ x² = 400
⇒ x = √400
⇒ x = ± 20
Reject x = – 20
[∵ Breadth cannot be negative.]
∴ x = 20
Hence, the rectangular grove is possible and its breadth = 20 m
and length 2 × 20 = 40 m.

Question 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.
Solution :
Let the present age of first friend be x years.
Then present age of second friend = (20 – x) years
Four years ago age of first friend = (x – 4) years
Four years ago age of second friend = (20 – x – 4) years = (16 – x) years
According to question,
(x – 4)(16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ – x² + 20x – 64 – 48 = 0
⇒ – x² + 20x – 112 = 0
⇒ x² – 20x + 112 = 0
Here, a = 1, b = – 20, c = 112
∴ D = b² – 4ac
⇒ D = (- 20)² – 4 × 1 × 112
D = 400 – 448
⇒ D = – 48
∵ D < 0
∴ The equation has no real value of x. Hence, the situation is not possible.

Question 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:
Let the breadth of rectangular park be x m.
Perimeter of rectangular park= 2 (length + breadth)
⇒ 80 = 2(length + x)
⇒ \(\frac{80}{2}\) = length + x
⇒ 40 = length + x
⇒ length = (40 – x) m
Now, according to question,
Area of rectangular park = Length × Breadth
400 = (40 – x) × x
⇒ 400 = 40x – x²
⇒ x² – 40x + 400 = 0
Here, a = 1, b = – 40, c = 400
D = b² – 4ac
⇒ D = (- 40)² – 4 × 1 × 400
⇒ D = 1600 – 1600
⇒ D = 0
The equation has real and equal roots.
⇒ x = 20 and 20
Hence, the rectangular park is possible and its length = 20 m and breadth = 20 m.