# Class10 NCRT Quadratic Equations Exercise – 4.4 pdf || UP Board

# __Quadratic
Equations__

__Exercise
– 4.4__

__Quadratic Equations__

__Exercise – 4.4__

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

** **

(i)
2x^{2} – 3x + 5 = 0

The given
equation is of the type ax^{2} + bx + c = 0, where a = 2, b = -3 and c
= 5

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

Hence no
real root exists.

(ii) 3x^{2}
- 4√3 x + 4 = 0

The given
equation is of the type ax^{2} + bx + c = 0, where a = 3, b = -4√3
and c = 4

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

Hence, the
given equation has real and equal roots.

Or
x = (2√3)/3, x = (2√3)/3

So the
roots of the given quadratic equation are (2√3)/3
and (2√3)/3.

(iii) 2x^{2}
– 6x + 3 = 0

The given
equation is of the type ax^{2} + bx + c = 0, where a = 2, b = -6
and
c =3

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

Hence, the
given equation has real and different roots.

Or
x = (3
+ √3)/2, x = (3
- √3)/2

So the
roots of the given quadratic equation are (3 + √3)/2 and (3 - √3)/2.

**2.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

The given
equation is of the type ax^{2} + bx + c = 0, where a = 2, b =
k
and c = 3

Therefore,
b^{2} – 4ac = (k)^{2} – 4 × 2 × 3 = k^{2} – 24

Since both
the roots of the equation are equal, so k^{2} – 24 = 0

k^{2}
= 24

k = ±√24

k = ±2√6

(ii) kx(x
– 2) + 6 = 0

kx^{2}
– 2kx + 6 = 0

The given
equation is of the type ax^{2} + bx + c = 0, where a = k, b
= -2k
and c = 6

Therefore,
b^{2} – 4ac = (-2k)^{2} – 4 × k × 6 = 4k^{2} – 24k

Since both
the roots of the equation are equal, so 4k^{2} – 24k = 0

4k(k – 6)
= 0

4k = 0, (k
– 6) = 0

k = 0, k =
6

But k ≠ 0
because it does not satisfy the equation kx(x – 2) + 6 = 0.

So, 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.**

Let the breadth
of the mango grove = x m

Hence,
length of the mango grove = 2x m

and area =
x × 2x = 2x^{2}

According
to Question,

2x^{2}
= 800

x^{2}
= 400

x = ±20

Since the breadth
of the mango grove cannot be negative, so the breadth of the mango grove =
20 m

and
length of the mango grove
= 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.**

Let the
age of the first friend = x years

Therefore,
age of second friend = 20 – x years

Four years
ago the age of the first friend = x – 4 years

Therefore,
age of the second friend four years ago = 20 – x – 4
= 16 – x years

According
to Question,

(x – 4)(16
– x) = 48

16x – x^{2}
– 64 + 4x = 48

x^{2}
– 20x + 112 = 0

The equation
is of the type ax^{2} + bx + c = 0
, where a = 1, b = -20
and c = 112

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

Hence no
real root exists.

Therefore,
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.**

Let the
length of the park b = x

Perimeter
= 80
m

Therefore,
breadth of the park = 40 – x

[Because
Perimeter = 2(Length + breadth)

80 = 2(x + breadth)

x + breadth
= 40

breadth = 40 – x]

According
to Question,

Area =
x(40 – x) = 400

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

Or
(x – 20)^{2} = 0

(x – 20) =
0

x = 20

Hence,
length of the park = 20
m

Therefore,
breadth of the park =
40 – x = 40 – 20 =
20 m

Download Class10 NCRT Quadratic Equations Exercise – 4.4 pdf

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