NCERT Solutions Class 10 Mathematics Quadratic Equations Exercise 4.4

Exercise 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

Answer :

(i) 2x² – 3x + 5 = 0                              
Comparing the given quadratic equation with ax² + bx + c = 0, we have:            
a = 2, b= – 3, c = 5           
Now discriminant = b² – 4ac            
                                 = (-3) ² – 4 * 2 * 5            
                                 = 9 – 40            
                                 = -31 < 0            
Since b² – 4ac is negative.          
Hence, the given quadratic equation has no real roots.
(ii) 3x² – 4√3x + 4 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we have:            
a = 3, b= -4√3, c = 4           
Now discriminant = b² – 4ac            
                                 = (-4√3) ² – 4* 3 * 4            
                                 = 48 – 48            
                                 = 0
Thus, the given quadratic equation has two real roots which are equal.
Here, the roots are: -b/2a and –b/2a
i.e -(-4√3)/(2 * 3) and -(-4√3)/(2 * 3)
⇒ 4√3/6 and 4√3/6
⇒ 2√3/3 and 2√3/3
⇒ 2√3/(√3 * √3) and 2√3/(√3 * √3)
⇒ 2/√3 and 2/√3                             
(iii) 2x² – 6x + 3 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we have:            
a = 2, b= -6, c = 3           
Now discriminant = b² – 4ac            
                                 = (-6) ² – 4* 2 * 3            
                                 = 36 – 24            
                                 = 12 > 0
Thus, the given quadratic equation has two real and distinct roots which are given as:
      x = {-b ± √(b² – 4 * a * c)}/2a
⇒ x = [-(-6) ± √{(-6)² – 4 * 2 * 3)}]/(2 * 2)
⇒ x = {6 ± √(36 – 24)}/4
⇒ x = (6 ± √12)/4
⇒ x = (6 ± 2√3)/4
⇒ x = 2(3 ± √3)/4
⇒ x = (3 ± √3)/2
Thus, the roots are: (3 + √3)/2 and (3 – √3)/2
 
 

Question : 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                                 (ii) kx(x – 2) + 6 = 0

Answer :

(i) 2x² + kx + 3 = 0                                
Comparing the given quadratic equation with ax² + bx + c = 0, we get            
a = 2, b = k, c = 3            
For a quadratic equation to have equal roots,            
     Discriminant = 0
⇒ b² – 4ac = 0
⇒ k² – 4 * 2 * 3 = 0
⇒ k² – 24 = 0
⇒ k² = 24
⇒ k = ±√24
⇒ k = ±2√6

(ii) kx(x – 2) + 6 = 0
⇒ kx² – 2kx + 6 = 0
Comparing the given quadratic equation with ax² + bx + c = 0, we get            
a = k, b = -2k, c = 6            
For a quadratic equation to have equal roots,            
     Discriminant = 0
⇒ b² – 4ac = 0
⇒ (-2k)² – 4 * k * 6 = 0
⇒ 4k² – 24k = 0
⇒ 4k(k – 6) = 0
⇒ k = 0, 6
But k cannot be 0 otherwise, the given equation is no more quadratic.
Thus, the required value of 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.

Answer :

Let the breadth be x metres.            
So, Length = 2x metres            
Now, Area = Length × Breadth            
                    = 2x * x             
                    = 2x² metre²            
According to the given condition,            
      2x² = 800
⇒ x² = 800/2
⇒ x² = 400
⇒ x = ±√400                        
⇒ x = 20 and x = -20            
But x = -20 is possible                                (Since breadth cannot be negative).            
So, x = 20            
Now 2x = 2 * 20 = 40            
Thus, length = 40 m and breadth = 20 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.

Answer :

Let the age of one friend = x years             
The age of the other friend = (20 – x) years             [Since sum of their ages is 20 years]            
Four years ago,            
Age of one friend = (x – 4) years            
Age of the other friend = (20 – x – 4) years       
                                          = (16 – x) years            
According to the condition,            
      (x – 4) × (16 – x) = 48          
⇒ 16x – 64 – x² – 4x = 48            
⇒ -x² – 20x – 64 – 48 = 0            
⇒ -x² – 20x – 112 = 0            
⇒ x² + 20x + 112 = 0   …………….1            
Here, a = 1, b = 20 and c = 112            
Now, b² – 4ac = (20) ² – 4 * 1 * 112            
                         = 400 – 448            
                         = – 48 < 0            
Since b² – 4ac is less than 0.            
So, the quadratic equation 1 has no real roots.            
Thus, the given equation is not possible.