**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.4 – Quadratic Equations. This Exercise contains 5 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 10 for Maths NCERT solutions for other Chapters, you can click the link at the end of this Note.**

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

**Solution:** We know that the quadratic equation ax^{2}+bx+c = 0 has

(a) Two distinct real roots, if b^{2}-4ac > 0,

(b) Two equal real roots, if b^{2}-4ac = 0,

(c) No real roots, if b^{2}-4ac < 0.

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

**Solution:** Comparing the given quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

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

Then, b^{2}-4ac = (-3)^{2}-4×2×5 = 9-40 = -31 < 0

Hence the given quadratic equation has no real roots.

**(ii) 3x ^{2}-4√3x+4 = 0**

**Solution:** Comparing the given quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

a = 3, b = -4√3 and c = 4

Then, b^{2}-4ac = (-4√3)^{2}-4×3×4 = 48-48 = 0

Hence the given quadratic equation has two equal real roots.

And the roots of the equation are given by

x =

x =

x =

x = and x =

x = and

Therefore, the root of the given quadratic equation are and

**(iii) 2x ^{2}-6x+3 = 0**

**Solution:** Comparing the given quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

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

then, b^{2}-4ac = (-6)^{2}-4×2×3 = 36-24 = 12 > 0

Hence the given quadratic equation has two distinct real roots.

And the roots of the equation are given by

x =

x =

x =

x =

x =

x = or x =

Therefore, the roots of the given quadratic equation are and

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

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

**Solution:** Comparing the given quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

a = 2, b = k and c = 3

Now the given quadratic equation have two equal roots if

b^{2}-4ac = 0

(k)^{2}-4×2×3 = 0

k^{2}-24 = 0

k^{2} = 24

k = ±

k = ± 2√6

Therefore, the required value of k is ± 2√6.

**(ii) kx(x-2)+6 = 0**

**Solution:** The given quadratic equation can be written as

kx^{2}-2kx+6 = 0….(1)

Comparing the quadratic equation (1) with the general form of quadratic equation ax^{2}+bx+c = 0 we get

a = k, b = -2k and c = 6

Now the given quadratic equation have two equal roots if

b^{2}-4ac = 0

(-2k)^{2}-4×k×6 = 0

4k^{2}-24k = 0

4k(k-6) = 0

k(k-6) = 0

k = 0

Or, k-6 = 0

k = 6

If k = 0, then equation will not have x^{2} and x, which is not possible because the given equation is quadratic equation.

Therefore, the required value of k is 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 mango grove be x m and the length is 2x m.

Area = length × breadth

= x × 2x

= 2x^{2} m^{2}

Then by the given condition,

2x^{2} = 800

x^{2} = 400

x = ±

x = ±20

Since, length cannot be negative, then x ≠ -20

Hence it is possible to design a rectangular mango grove whose length is twice its breadth and the area is 800 m^{2} and its breadth is 20 m and length is 20×2 = 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 1^{st} friend is x.

Then the age of 2^{nd} friend is (20-x)

Four year ago their age was (x-4) and (20-x-4)

Then using the given condition we have

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

16x – 64 – x^{2} +4x = 48

20x – x^{2 }– 64 – 48 = 0

x^{2 }– 20x + 112 = 0……(1)

Now comparing the above quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

a = 1, b = -20 and c = 112

then, b^{2}-4ac = (-20)^{2}-4×1×112 = 400-448 = -48 > 0

Therefore, no real root is possible for this equation and hence this 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 and breadth of the park be “l” and “b”

Area of rectangle = 2(l + b)

2(l + b) = 80

l + b = 40

b = 40 – l

Then, Area = l(40 – l)

Then by the given condition,

l(40 – l) = 400

40l – l^{2} -400 = 0

l^{2} – 40l + 400 = 0….(1)

Now comparing the above quadratic equation with the general form of quadratic equation ax^{2}+bx+c = 0 we get

a = 1, b = -40 and c = 400

Then, b^{2}-4ac = (-40)^{2}-4×1×400 = 1600-1600 = 0

As the quadratic equation has two equal roots then the given situation is possible.

And the solution of the equation (1) is given by

x =

x =

x =

x = = 20

Therefore, the length of the park is 20 m and the breadth of the park = 40 – l = 40-20 = 20 m.

**NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.4 – Quadratic Equations, has been designed by the NCERT to test the knowledge of the student on the topic – Nature of Roots**

**Download NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.4 – Quadratic Equations**

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