NCERT Solutions for Quadratic Equations Chapter 4 Ex 4.3 Class 10 Maths has all the solutions to the questions provided in the NCERT Book of the latest edition.
Students are advised to practice all the questions to get good marks in the board examination.
| Textbook | NCERT |
| Class | 10 |
| Subject | Mathematics |
| Chapter | 4 |
| Exercise | 4.3 |
| Chapter Name | Quadratic Equations |
Class 10 Maths Chapter 4 Exercise 4.3 Quadratic Equations NCERT Solution
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) 2x^{2}– 3x + 5 = 0
Answer
2x^{2}-3x+5=0
\therefore D = b^{2}-4ac
=(-3)^{2}-4\times 2 \times 5
=9-40 = -31
\because D < 0
(ii) 3x^{2}– 4\sqrt{3}x + 4 = 0
Answer
3x^{2}-4\sqrt{3}x+4=0
\therefore D=(-4\sqrt{3})^{2}-4\times 3 \times 4
=48-48=0
\because D=0
(iii) 2x^{2} – 6x + 3 = 0
Answer
2x^{2}-6x+3=0
\therefore D=(-6)^{2}-4\times 2 \times 3
=36-24=12
\therefore
Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2x^{2}+ kx + 3 = 0
Answer
2x^{2}+kx+3=0
\because
\therefore D = 0
⇒ b^{2}-4ac=0
⇒ K^{2}-4\times2\times3=0
⇒ K^{2}=24
⇒ K=2\sqrt{6}
(ii) kx (x – 2) + 6 = 0
Answer
kx(x-2)+6=0
⇒ kx^{2}-2kx+6=0
\therefore D = 0
⇒ 4k^{2}-24k=0
⇒ 4k(k-6)=0
\therefore k=0
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m^{2}
Answer
Let breadth of groove be x m
then length of groove = 2x m
\therefore
⇒ 2x\times x=800
⇒ x^{2}=400
⇒ x=\pm20
\because
So, Breadth = x = 20m
length = 2x = 40m
\therefore
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 present age of 1^{st}
then the present age of 2^{nd}
A/Q,
4 year ago
1^{st}
2^{nd}
\therefore (x-4)(16-x)=48
⇒ 16x-x^{2}-64+4x=48
⇒ 20x-x^{2}=48+64
⇒ x^{2}-20x+112=0
D=(20)^{2}-4\times 12
=400-448 = -48
\because D<0
\therefore
Is it possible to design a rectangular park of perimeter 80 m and area 400 m^{2}
Answer
Perimeter of rectangle = 2(l+b)
Area of rectangle = length\times breadth
Let the length of park be x
then the breadth of park = \frac{80}{2}-x
According to question,
x\times (40-x) = 400m^{2}
⇒ 40x-x^{2}=400
⇒ x^{2}-40x+400=0
⇒ x^{2}-20x-20x+400=0
⇒ x(x-20)-20(x-20)=0
⇒ (x-20)(x-20) = 0
⇒ x-20=0
⇒ x=20
\therefore
breadth of park = 40 – 20 = 20m
\therefore
Hope NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.3, helps you in solving problems. If you have any doubts, drop a comment below and we will get back to you.