Quadratic Equation

NCERT Exercise 4.4

Question (1) Find the nature of the roots of the following equations. If the real roots exist, find them

Question (1)(i) 2x²– 3x + 5

Solution:

Given, 2x² – 3x + 5

Here, a = 2, b = – 3 and c = 5

We know that D = b² – 4ac

∴ D =(– 3)² – 4 × 2 × 5

⇒ D = 9 – 40 = – 31

Here, since, D<0

∴ Real roots for the given quadratic equation do not exist

Question (1)(ii)3x² – 4 √3 x + 4 = 0

Solution:

Given, 3x² – 4 √3 x + 4 = 0

Here, a = 3, b = – 4 √ 3 and c = 4

We know that D = b² – 4ac

∴ D =(– 4√ 3 )^{– 4} × 3 × 4

⇒ D = 16 × 3 – 48

⇒ D = 48 – 48 = 0

Here, since D = 0, therefore roots for the given quadratic equation are real and equal

Now, we know that roots =^{– b ± √ D}/_2a

= ^{(– (– 4 √3)± √ 0} /_{2 × 3}

= ^{4 √ 3}/₆

⇒ Root = ^{2 √ 3} /₃ Answer

Question(1)(iii)2x² – 6x + 3 = 0

Solution:

Given, 2x² – 6x + 3 = 0

Here, a = 2, b = – 6 and c = 3

We know that D = b² – 4ac

∴ D = (– 6)² – 4 × 2 × 3

⇒ D = 36 – 24

⇒ D = 12

∵ D > 0, ∴ given quadratic equation has two real roots.

Now, we know that roots = ^{– b ± √ D}/_2a

∴ Roots = ^{– ( – 6) ± √ 12}/_{2 × 2}

= ^{6 ± √3 × 4}/₄

= ^{6 ± 2√3}/₄

=^{2(3 ± √ 3)}/₄

= ^{3 ± √ 3} /₂

∴ roots = ^{3 + √3}/₂ and ^{3 – √3}/₂ Answer

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

Question (2)(i) 2x² + kx + 3 = 0

Solution:

We know that, D = b² – 4ac and for equal roots a quadratic equation must have D = 0

Given,2x² + kx + 3 = 0

Here, a = 2, b = k and c = 0

∴ D = k² – 4 × 2 × 3

⇒ k² – 24 = 0

[∵ for equal roots D = 0]

⇒ k² = 24

⇒ k = √24

⇒ k = √6 × 4

⇒ k = 2 √6Answer

Question (2)(ii) kx(x – 2) + 6 = 0

Solution:

Given, kx(x – 2) + 6 = 0

⇒ kx² – 2kx + 6 = 0

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

We know that, D = b² – 4ac

∴ D =(2k)² – 4 × k × 6

We know that, for equal roots a quadratic equation must have D = 0

Thus, by putting the value of D = 0 we get

(2k)² – 4 × k × 6 = 0

⇒ 4k² – 24k = 0

⇒ 4(k² – 6k) = 0

⇒ k² – 6k = 0

⇒ k² = 6k

⇒ k = 6 Answer

Question (3) Is it possible to design a rectangular mango grove whose lenght is twice its breadht, and the area is 800 m²? If so, find its length and breadth.

Solution:

Let, breadth of the grove = x

∴ according to question, lenght = 2x

Given, Area = 800 m²

We know that, Area of a rectangle = lenght x breadth

∴ 800m² = 2x × x

⇒ 2x² = 800m²

⇒ 2x² – 800 = 0

⇒ 2(x² – 400) = 0

⇒ x² – 400 = 0

Here, a = 1, b = 0 and c = 400

We know that if D > 0, then only roots are possible

We know that, D = b² – 4ac

⇒ D = 0² – 4 × 1 × 400

⇒ D = 1600

Since, D > 0, thus roots for the given quadratic equation is possible. And hence, a reactangular mango grove in given condition in quesion is possible to design.

Now, roots x = ^{– b2 ±√D}/_{2 a}

x = ^{– (0)2 ±√1600}/_{2 ×1}

x =(± 40)2

x = ± 20

∴ x = 20 or – 20

since, breadht cannot be negative, thus, by discarding negative value, we have

Lenght = 40m & breadth = 20m Answer

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:

Given, the sum of the age of two friends = 20 years

Again given, four years ago, the product of their age = 48 years

Let Age of Friend (A) = a year

Age of friend (a) 4 years ago = a – 4 year

∴ Age of Friend (B) = 20 – a year

And ∴ Age of friend (B) 4 yers ago = 20 – a – 4 = 16 – a year

Thus, According to question:

(a – 4)(16 – a) = 48

⇒ 16a – a² – (16 × 4) + 4a = 48

⇒ 16a – a² – 64 + 4a = 48

⇒ – a² + 16a + 4a – 64 = 48

⇒ – a² + 20a – 64 – 48 = 0

⇒ – a² + 20a – 112 = 0

By taking minus as common, we get

– (a² – 20a + 112) = 0

⇒ a² – 20a + 112 = 0 ----(i)

Now, to know whether, root of the above quadratic equation (i) exits or not, we need to know the value of D

We know that, D = b² – 4ac

Here, in equation (i) a = 1, b = – 20 and c = 112

∴ D = (– 20)² – 4 × 1 × 112

⇒D = 400 – 448 = – 48

Here, since, D < 0, thus roots of the given equation (i) is not possible.

Hence, situation given in the question is not possible.

Question (5) Is it possible to design a rectangular park of perimeter 80m and area 400m²? If so find its length and breadth.

Solution:

Given, perimeter of park = 80m

And Area of the park = 400m²

Let, Length of the given rectangular park = l

And Breadth of the given rectangular park = b

We know that, Perimeter of a rectanle = 2(l + b)

∴ 80 m = 2(l + b)

⇒ l + b = ⁸⁰/₂

⇒ l + b = 40

⇒ l = 40 – b

Now, we know that, Area of a rectanlge = l × b

∴ 400 = l × b

By putting the value of l = 40 – b we get

400 = (40 – b) × b

⇒ 40b – b² = 400

⇒ – b² + 40b – 400 = 0

By taking minus as common, we get

– (b² – 40b + 400) = 0

⇒ b² – 40b + 400 = 0 ----(i)

Now, to see whether root of the equation (i) exists or not, we need to calculate the value of D

We know, that, D = b² – 4ac ----(ii)

In equation (i)

a = 1, b = – 40 and c = 400

By putting values of a,b and c in equation (i), we get

D = (– 40)² – 4× 1 × 400

⇒ D = 1600 – 1600 = 0

Here, since, D = 0, thus equal roots are possible for the given equation (i), hence a rectangular park is possible to design according to condition given in the question

We know that, Roots = ^{– b ±√D}/_{2 a}

∴ Roots = ^{– (– 40) ±√0}/_{2 × 1}

⇒ Roots = ⁴⁰/₂ = 20

Thus, Breadth (b) of the given, park = 20m.

And, Lenght (l) = 40 – b = 40 – 20 = 20m

Thus, sides of the given park = 20m Answer

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