In this post, you will find the NCERT solutions for class 10 maths ex 4.3. These solutions are based on the latest syllabus of NCERT Maths class 10.

**NCERT Solutions for Class 10 Maths Ex 4.3**

**1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:**

**(i) 2 x^{2 }**

**− 3**

*x*+ 5 = 0**(ii) 3 x^{2 }**

**− 4√3**

*x*+ 4 = 0**(iii) 2 x^{2 }**

**− 6**

*x*+ 3 = 0** **

**Solution: (i) 2 x^{2 }**

**− 3**

*x*+ 5 = 0Comparing the equation 2*x*^{2 }− 3*x* + 5 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 2, *b *= −3 and *c *= 5

Discriminant (D) = *b*^{2 }− 4*ac* = (−3)^{2 }– 4 × 2 × 5

= 9 – 40 = −31

Here, D < 0, which means that the equation has no real roots.

** **

**(ii) 3 x^{2 }**

**− 4√3**

*x*+ 4 = 0Comparing the equation 3*x*^{2 }− 4√3*x* + 4 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 3, *b *= −4√3 and c = 4

Discriminant (D) = *b*^{2 }− 4*ac* = (−4√3)^{2 }− 4 × 3 × 4

= 48 – 48 = 0

Here, D = 0, which means that the equation has two equal real roots.

Using the quadratic formula, we have

*a*,

*b*and

*c*, we get

Since the equation has two equal roots, it means

**(iii) 2 x^{2 }**

**− 6**

*x*+ 3 = 0Comparing the equation 2*x*^{2 }− 6*x* + 3 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 2, *b *= −6 and *c *= 3

Discriminant (D) = *b*^{2 }− 4*ac* = (−6)^{2 }− 4 × 2 × 3

= 36 – 24 = 12

Here, D > 0, which means that the equation has two distinct real roots.

Using the quadratic formula, we have

*a*,

*b*and

*c*, we get

**2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.**

**(i) 2 x^{2 }+ kx + 3 = 0**

**(ii) kx(x − 2) + 6 = 0**

** **

**Solution: (i) 2 x^{2 }+ kx + 3 = 0**

We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.

Comparing the equation 2*x*^{2 }+ *kx* + 3 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 2, *b *= *k* and *c *= 3

Discriminant (D) = *b*^{2 }− 4*ac* = *k*^{2 }– 4 × 2 × 3 = *k*^{2 }− 24

Since the equation has two equal roots, we have

D = 0

*k*^{2 }− 24 = 0

*k*^{2} = 24

⇒ *k* = ±√24 = ±2√6

⇒ *k* = 2√6, −2√6

** **

**(ii) kx(x − 2) + 6 = 0**

⇒ *kx*^{2} – 2*kx* + 6 = 0

Comparing the equation *kx*^{2} – 2*kx* + 6 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= *k*, *b *= −2*k* and *c *= 6

Discriminant (D) = *b*^{2 }− 4*ac* = (−2*k*)^{2 }− 4 × *k *× 6 = 4*k*^{2 }− 24*k*

We know that a quadratic equation has two equal real roots, if the value of discriminant (D) is equal to zero.

Since the equation has two equal roots, we have

D = 0

4*k*^{2 }− 24*k* = 0

⇒ 4*k*(*k *− 6) = 0

⇒ *k *= 0, 6

According to the definition of a quadratic equation, we have

A quadratic equation is the equation of the form *ax*^{2 }+ *bx* + *c* = 0, where *a *≠ 0.

Therefore, in equation *kx*^{2 }− 2*kx* + 6 = 0, we cannot have *k* = 0.

Therefore, we discard *k* = 0.

Hence, the answer is *k* = 6.

** **

**3. 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.**

** **

**Solution: **Let the breadth of the rectangular mango grove be *x* m.

Then, the length of the rectangular mango grove = 2*x* m

Area of the rectangular mango grove = length × breadth= *x* × 2*x* = 2*x*^{2 }m^{2}

According to the question, we have

2*x*^{2} = 800

⇒ 2*x*^{2 }− 800 = 0

⇒ *x*^{2 }− 400 = 0

Comparing the equation *x*^{2 }− 400 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 1, *b *= 0 and *c *= −400

Discriminant (D) = *b*^{2 }− 4*ac* = (0)^{2 }− 4 × 1 × (−400) = 1600

Since D > 0, it means that the equation has two distinct real roots.

Therefore, it is possible to design a rectangular mango grove.

Using the quadratic formula, we have

Putting the values of *a*, *b* and *c*, we get

⇒ *x *= 20, −20

We discard the negative value of *x* because the breadth of a rectangle cannot be negative.

Therefore, *x* = breadth of the rectangle = 20 m

And, length of the rectangle = 2*x *= 2 × 20 = 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.**

** **

**Solution: **Let the age of first friend be *x* years.

Then, the age of second friend = (20 − *x*) years.

Four years ago, the age of first friend = (*x *− 4) years

Four years ago, the age of second friend = (20 − *x*) − 4 = (16 − *x*) years

According to the question, we have

(*x *− 4)(16 − *x*) = 48

⇒ 16*x *− *x*^{2 }– 64 + 4*x* = 48

⇒ 20*x* − *x*^{2 }− 112 = 0

⇒ *x*^{2 }− 20*x* + 112 = 0

Comparing the equation *x*^{2 }− 20*x* + 112 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 1, *b *= −20 and *c *= 112

Discriminant (D) = *b*^{2 }− 4*ac* = (−20)^{2 }– 4 × 1 × 112 = 400 – 448 = −48

Here D < 0, which means we have no real roots for this equation.

Therefore, the give situation is not possible.

** **

**5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m ^{2}? If so, find its length and breadth.**

** **

**Solution: **Let the length of the park be *x* m.

It is given that the area of the rectangular park = 400 m^{2}

Length × breadth = 400 m^{2}

Therefore, breadth of park = 400/*x* m

Perimeter of the rectangular park = 2(length + breath) = 2(*x* + 400/*x*) m

It is given that the perimeter of the rectangular park = 80 m

According to the question, we have

⇒ 2*x*^{2 }+ 800 = 80*x*

⇒ 2*x*^{2 }− 80*x* + 800 = 0

⇒ *x*^{2 }− 40*x* + 400 = 0

Comparing the equation *x*^{2 }− 40*x* + 400 = 0 with the general form of a quadratic equation *ax*^{2 }+ *bx* + *c* = 0, we get

*a *= 1, *b *= −40 and *c *= 400

Discriminant (D) = *b*^{2 }− 4*ac* = (−40)^{2 }− 4 × 1 × 400 = 1600 – 1600 = 0

Here D = 0.^{}

Therefore, the two roots of the equation are real and equal, which means that it is possible to design a rectangular park of the perimeter 80 m and area 400 m^{2}.

Using the quadratic formula, we have

Putting the values of *a*, *b* and *c*, we get

Here, both the roots are equal to 20.

Therefore, the length of the rectangular park = 20 m

And, the breadth of the rectangular park = 400/*x* = 400/20 = 20 m