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
$$2x^{2}-3x+5=0$$
$$\therefore D = b^{2}-4ac$$
$$=(-3)^{2}-4\times 2 \times 5$$
$$=9-40 = -31$$
$$\because D < 0$$, So the equation has no real roots
(ii) $$3x^{2}– 4\sqrt{3}x + 4 = 0$$
Answer
$$3x^{2}-4\sqrt{3}x+4=0$$
$$\therefore D=(-4\sqrt{3})^{2}-4\times 3 \times 4$$
$$=48-48=0$$
$$\because D=0$$, So the equation has two equal real roots.
(iii) $$2x^{2} – 6x + 3 = 0$$
Answer
$$2x^{2}-6x+3=0$$
$$\therefore D=(-6)^{2}-4\times 2 \times 3$$
$$=36-24=12$$
$$\therefore$$ D > 0, So the equation has two distinct real roots.
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
$$2x^{2}+kx+3=0$$
$$\because$$ Roots of equation are equal
$$\therefore D = 0$$
⇒ $$b^{2}-4ac=0$$
⇒ $$K^{2}-4\times2\times3=0$$
⇒ $$K^{2}=24$$
⇒ $$K=2\sqrt{6}$$
(ii) $$kx (x – 2) + 6 = 0$$
Answer
$$kx(x-2)+6=0$$
⇒ $$kx^{2}-2kx+6=0$$
$$\therefore D = 0$$
⇒ $$4k^{2}-24k=0$$
⇒ $$4k(k-6)=0$$
$$\therefore k=0$$ and $$k=6$$
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 $$m^{2}$$? If so, find its length and breadth.
Answer
Let breadth of groove be x m
then length of groove = 2x m
$$\therefore$$ Area of groove = 800 $$m^{2}$$
⇒ $$2x\times x=800$$
⇒ $$x^{2}=400$$
⇒ $$x=\pm20$$
$$\because$$ Breadth can’t be negative
So, Breadth = x = 20m
length = 2x = 40m
$$\therefore$$ Yes, it is possible to design a rectangular many groove
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 $$1^{st}$$ friend Be x years
then the present age of $$2^{nd}$$ friend = $$(20-x)$$ years
A/Q,
4 year ago
$$1^{st}$$ friend = $$x-4$$
$$2^{nd}$$ friend age = $$(20 – x – 4)$$ = $$16 – x$$
$$\therefore (x-4)(16-x)=48$$
⇒ $$16x-x^{2}-64+4x=48$$
⇒ $$20x-x^{2}=48+64$$
⇒ $$x^{2}-20x+112=0$$
$$D=(20)^{2}-4\times 12$$
$$=400-448 = -48$$
$$\because D<0$$, the roots of equation are not real
$$\therefore$$ The following situation is not possible
Is it possible to design a rectangular park of perimeter 80 m and area 400 $$m^{2}$$? If so, find its length and breadth.
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 = $$\frac{80}{2}-x$$ = $$40-x$$
According to question,
$$x\times (40-x) = 400m^{2}$$
⇒ $$40x-x^{2}=400$$
⇒ $$x^{2}-40x+400=0$$
⇒ $$x^{2}-20x-20x+400=0$$
⇒ $$x(x-20)-20(x-20)=0$$
⇒ $$(x-20)(x-20) = 0$$
⇒ $$x-20=0$$
⇒ $$x=20$$
$$\therefore$$ length of park = x = 20m
breadth of park = 40 – 20 = 20m
$$\therefore$$ Yes, this design is possible
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.