NCERT Class 10th Math chapter 4 Exercise 4.4 Question Answer. NCERT Solution for Class 10 Maths Chapter 4 Exercise 4.4 of Quadratic Equations NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.
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NCERT Class 10 Math chapter 4 Quadratic Equations Exercise 4.4 Question Answer Solution.
Class 10 Math Chapter 4 Ex. 4.4 Question Answer
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0
Ans. Checking roots of quadratic equation,
b2 – 4ac = (-3)2 – 4 × 2 × 5 = 9 – 40 = – 31
As b2 – 4ac < 0, no real roots exist.
(ii) 3x2 – 4√3x + 4 = 0
Ans. Checking roots of quadratic equation,
b2 – 4ac = (-4√3)2 – 4 × 3 × 4 = 48 – 48 = 0
As b2 – 4ac = 0, equal real roots exist.
Now finding that roots.
3x2 – 4√3x + 4 = 0
Using quadratic formula to find roots.
Formula : {-b ± √(b2 – 4ac)} / 2a
= {4√3 ± √0} / 2(3)
= 4√3/6
= 2√3/3
As roots are equal. Therefore, roots of given quadratic equation are (2√3) / 3 and (2√3) / 3
(iii) 2x2 – 6x + 3 = 0
Ans. Checking roots of quadratic equation,
b2 – 4ac = (-6)2 – 4(2)(3) = 12
As b2 – 4ac > 0, Distinct real roots exist.
Now finding that roots.
2x2 – 6x + 3 = 0
We cannot factorise them. So we are using quadratic formula to find roots.
Formula : {-b ± √(b2 – 4ac)} / 2a
= {6 ± √12} / 4
= {3 ± √3} / 2
2. Find the vales of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x2 + kx + 3 = 0
Ans. To get equal roots, we have to make b2 – 4ac = 0
On comparing to quadratic equation,
k2 – 4(2)(3) = 0
k2 – 48 = 0
k2 = 48
k = ± √48
k = ± 2√6
(ii) kx(x – 2) + 6 = 0
Ans. kx2 -2kx + 6 = 0
To get equal roots, we have to make b2 – 4ac = 0
On comparing to quadratic equation,
(-2k)2 – 4(k)(6) = 0
4k2 – 24k = 0
4k2 = 24k
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.
Ans. Let x ( in metre) be the breadth of given rectangular mango grove.
Then 2x will be the length of given rectangular mango grove.
According to question,
x(2x) = 800
2x2 = 800
x2 = 400
x = ± 20
As distance cannot be negative. Therefore, taking positive value only.
We find breadth of rectangle be 20 m and length be 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.
Ans. Let x be the the age of first friend.
Then 20 – x will be the age of second friend.
Four years ago product of their ages was 48.
(x – 4)(20 – x – 4) = 48
(x – 4)( 16 – x) = 48
16x – x2 -64 + 4x= 48
– x2 + 20x – 64 – 48 = 0 (rearranging)
x2 – 20x + 112 = 0
As b2 – 4ac = (-20)2 – 4(1)(112) = 448
b2 – 4ac < 0, no real roots exist.
Therefore, the following situation is not possible.
5. Is it possible to design a rectangular park of perimeter 80m and area 400m2 ? If so, find its length and breadth.
Ans. As we know that perimeter of any rectangle = 2(length + breadth) = 80
Length + Breadth = 40
Now, let x be the length of rectangle. Then 40 – x will be breadth of rectangle.
Area of rectangle = 400
x(40 – x) = 400
40x – x2 – 400 = 0 ( rearranging )
x2 – 40x + 400 = 0
As b2 – 4ac = (-40)2 – 4(1)(400) = 0
b2 – 4ac = 0, equal real roots exist.
Now finding the roots.
x2 – 40x + 400 = 0
x2 – 20x – 20x + 400 = 0
x(x – 20) – 20(x – 20) = 0
(x – 20)(x – 20) = 0
x = 20 or x = 20
Therefore, Length is 20m and breadth is (40 – 20) which is 20m also .