NCERT Solutions for Class 10 Maths Chapter 4 – Quadratic Equation: Exercise 4

This exercise contains problems based on the nature of the root of the quadratic equation which can be determined using the quadratic formula. Other than that same word problems are based on real-world scenarios.

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

Solution:

(i) Given: 2x²-3x+5=0

Here a=2,b=-3 and c=5

[Tex]\therefore         [/Tex] Discriminant, D=b²-4ac

⇒ D = (-3)²– 4 × 2 × 5)

⇒ D = 9-40 = -31 < 0

Hence, the roots are imaginary.

(ii) 3x²-4√3x+4=0

Solution:

(ii) Given: 3x²-4√3x + 4 = 0

Here a=3,b=√3 and c=4

[Tex]\therefore         [/Tex] Discriminant, D=b²-4ac

⇒ D = (-4√3)² – (4 × 3 × 4)

⇒ D= 48 – 48 = 0

Hence, the roots are real and equal.

Using the formula,

[Tex]x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}                [/Tex], we get 

[Tex]x=\frac {-(-4\sqrt3)\pm\sqrt{(-4\sqrt3)^2-4\times3\times4}} {2\times3} [/Tex]

[Tex]= \frac {4\sqrt3\pm\sqrt{48-48}} {6}=\frac {4\sqrt3} {6}=\frac {2} {\sqrt3} [/Tex]

Hence, the equal roots are [Tex]\frac 2 {\sqrt3}         [/Tex] and [Tex]\frac 2 {\sqrt3}                [/Tex].

(iii) 2x²-6x+3=0

Solution:

(iii) Given: 2x²-6x+3=0

Here, a=2,b=-6 and c=3

[Tex]\therefore         [/Tex] Discriminant, D=b²-4ac

⇒ D = (-6)² – (4 × 2 × 3)

⇒ D = 36 – 24 = 12 > 0

Hence, the roots are distinct and real.

Using the formula,

[Tex]x=\frac {-b\pm\sqrt{b^2-4ac}} {2a}                [/Tex],we get

[Tex]x=\frac {-(-6)\pm\sqrt{(-6)^2-4\times2\times3}} {2\times2}  [/Tex]

[Tex]x=\frac {6\pm\sqrt{36-24}} {4}  [/Tex]

[Tex]x=\frac {6\pm\sqrt{12}} {4}  [/Tex]

[Tex]x=\frac {6\pm2\sqrt{3}} {4}  [/Tex]

[Tex]x=\frac {3\pm\sqrt{3}} {2}  [/Tex]

Hence, the equal roots are [Tex]\frac {3+\sqrt{3}} {2}         [/Tex]and [Tex]\frac {3-\sqrt{3}} {2}  [/Tex]

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

Solution:

This equation is of the form ax²+bx+x, where a=2, b=k and c=3.

Discriminant, D=b²-4ac

⇒D= k²  – 4 × 2 × 3

⇒D = k² -24

For equal roots D=0

⇒ k²-24=0

⇒ k²=24

⇒ k² = ±24 = ±2√6​
 

(ii) kx(x-2)+6=0

Solution:

⇒  kx²-2kx+6=0

This equation is of the form ax²+bx+c=0, where a=k, b=-2k and c=6.

Discriminant, D=b²-4ac

⇒ D =(-2k)² – 4 × k × 6

⇒ D =4k²-24k

For equal roots D=0

⇒  4k²-24k=0

⇒  4k(k-24)=0

⇒ k=0 (not possible) or 4k-24=0

⇒  k= 24/4=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 mango grove be x m.

Then, the length of the rectangular mango grove will be 2x m.

The Area of the rectangular mango grove=length × breadth

According to the question, we have

x × 2x= 800

⇒  2x²=800

⇒  x²=400

⇒  x=20

Hence, the rectangular mango grove is possible to design whose 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.

Solution:

Let the present age of one friend be x years.

Then, the present age of other friend be (20-x) years.

4 years ago, one friend’s age was (x-4) years

4 years ago, other friend’s age was (20-x-4)=(16-x) years.

According to the question,

(x-4)(16-x)=48

⇒  16x-64-x²+4x=48

⇒  x²-20x+112=0

This equation is of the form ax²+bx+c=0,where a=1, b=-20 and c=112.

Discriminant, D=b²-4ac

⇒ D = (-20)²-4 × 1 × 112 = -48 < 0

Since, there are no real roots.

So the given 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 length of the rectangular park be x.

The perimeter of the rectangular park= 2(length + breadth)

⇒  2(x + breadth)=80

⇒  breadth=40-x

The area of rectangular park= length × breadth

⇒  x(40-x)=400

⇒  40x-x²=400

⇒  x²-40x+400=0

⇒  x² -20x-20x+400=0

⇒ (x-20)(x-20)=0

⇒ x=20

Hence, the rectangular park is possible to design. So, the length of the park is 20m and the breadth = 40-20=20m.

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations is a resource created by the team at GFG to help students clear their doubts while solving problems from the NCERT textbook. It helps clear the frustration caused by being stuck on a problem for a while. The NCERT Solutions for Class 10 Maths covers all the questions in the exercise of the NCERT textbook for this chapter.

In this NCERT Class 10 Maths Chapter 4 Solutions, students are introduced to the world of quadratic equations, although quadratic equations are introduced to students in Class 9 under the chapter named “Polynomials” there was no separation of this from polynomials but in reality, there are many applications of quadratic equation, therefore it is studied here in class 10 maths as a stand-alone chapter. In this chapter, students learn about quadratic equations, various methods to solve quadratic equations, the nature of roots based on discriminants for quadratic equations, and many real-life based problems which can be modelled in the form of quadratic equations and then solved to get the required solution.