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

**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.

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

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