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

(i) 2x^{2} âˆ’ 3x + 5 = 0

(ii) 3x^{2} âˆ’ 4âˆš3 x+ 4 = 0

(iii) 2x^{2} âˆ’ 6x + 3 = 0

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

Comparing the given quadratic equation with ax^{2} + bx + c = 0, we have:

a = 2

b = âˆ’ 3

c = 5

âˆ´ The discriminant = b^{2} âˆ’ 4ac

= (âˆ’ 3)^{2 }âˆ’ 4 (2) (5)

= 9 âˆ’ 40

= âˆ’ 31 < 0

Since b^{2} âˆ’ 4ac is negative.

âˆ´ The given quadratic equation has no real roots.

(ii) 3x^{2} - 4âˆš3x + 4 = 0

Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get

a = 3

b = âˆ’ 4âˆš3

c = 4

âˆ´

= (16 Ã— 3) âˆ’ 48

= 48 âˆ’ 48 = 0

Thus, the given quadratic equation has two real roots which are equal. Here, the roots are:

â‡’

Thus,

(iii) 2x^{2}âˆ’6x + 3 = 0

Comparing it with the general quadratic equation, we have:

a = 2

b = âˆ’ 6

c = 3

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 6)^{2} âˆ’ 4 (2) (3)

= 36 âˆ’ 24

= 12 > 0

âˆ´ The given quadratic equation has two real and distinct roots, which are given by

â‡’

Thus, the roots are:**Ques ****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

(ii) kx (x âˆ’ 2) + 6 = 0**Sol:** (i) 2x^{2} + kx + 3 = 0

Comparing the given quadratic equation with ax^{2} + bx + c = 0, we get

a = 2

b = k

c = 3

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ k)^{2} âˆ’ 4 (2) (3)

= k^{2} âˆ’ 24

âˆµ For a quadratic equation to have equal roots,

b^{2 }âˆ’ 4ac = 0

âˆ´ k^{2} âˆ’ 24 = 0 â‡’ k =

â‡’

Thus, the required values of k are

(ii) kx (xâˆ’2) + 6 = 0

Comparing kx (x âˆ’ 2) + 6 = 0 i.e., kx^{2} âˆ’ 2kx + 6 = 0 with ax^{2 }+ bx + c = 0, we get

a = k

b = âˆ’ 2k

c = 6

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 2k)^{2} âˆ’ 4 (k) (6)

= 4k^{2} âˆ’ 24k

Since, the roots are real and equal,

âˆ´ b^{2} âˆ’ 4ac =0

â‡’ 4k^{2} âˆ’ 24k =0

â‡’ 4k (k âˆ’ 6) = 0

â‡’ 4k = 0 or k âˆ’ 6 = 0

â‡’ k = 0 or k = 6

But k cannot be 0, otherwise, the given equation is no more quadratic. Thus, the required value of k = 6.**Ques ****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.**

âˆ´ Length = 2x metres

Now, Area = Length Ã— Breadth

= 2x Ã— x metre2

= 2x

According to the given condition,

2x^{2} = 800

â‡’

â‡’

Therefore, x = 20 and x = âˆ’ 20

But x = âˆ’ 20 is possible |(âˆµ breadth cannot be negative).

âˆ´ x = 20

â‡’ 2 x = 2 Ã— 20 = 40

Thus, length = 40 m and breadth = 20 m**Ques ****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.****Sol:** Let the age of one friend = x years

âˆ´ The age of the other friend = (20 âˆ’ x) years [âˆµ Sum of their ages is 20 years]

**Four years ago Age **

of one friend = (x âˆ’ 4) years

Age of the other friend = (20 âˆ’ x âˆ’ 4) years

= (16 âˆ’ x) years

According to the condition,

(x âˆ’ 4) Ã— (16 âˆ’ x) = 48

â‡’ 16x âˆ’ 64 âˆ’ x^{2} âˆ’ 4x = 48

â‡’âˆ’ x^{2} âˆ’ 20x âˆ’ 64 âˆ’ 48 = 0

â‡’âˆ’ x^{2} âˆ’ 20x âˆ’ 112 = 0

â‡’ x^{2} + 20x + 112 = 0 ....(1)

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

âˆ´ b^{2 }âˆ’ 4ac = (20)^{2} âˆ’ 4 (1) (112)

= 400 âˆ’ 448

= âˆ’ 48 < 0

Since b^{2} âˆ’ 4ac is less than 0.

âˆ´ The quadratic equation (1) has no real roots.

Thus, the given equation is not possible.**Ques ****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.**

**Sol: **Let the breadth of the rectangle be x metres.

Since, the perimeter of the rectangle = 80 m.

âˆ´ 2 [Length + Breadth] = 80

2 [Length + x] = 80

â‡’

â‡’ Length = (40 âˆ’ x) metres

âˆ´ Area of the rectangle = Length Ã— breadth

= (40 âˆ’ x) Ã— x sq. m

= 40x âˆ’ x^{2}

Now, according to the given condition,

Area of the rectangle = 400 m^{2}

âˆ´ 40x âˆ’ x^{2} = 400

â‡’ âˆ’ x^{2} + 40x âˆ’ 400 = 0

â‡’ x^{2} âˆ’ 40x + 400 = 0 ...(1)

Comparing (1) with ax^{2} + bx + c = 0, we get

a = 1

b = âˆ’ 40

c = 400

âˆ´ b^{2} âˆ’ 4ac =(âˆ’ 40)^{2} âˆ’ 4 (1) (400)

= 1600 âˆ’ 1600 = 0

Thus, the equation (1) has two equal and real roots.

âˆµ

âˆ´

âˆ´ Breadth, x = 20 m

âˆ´ Length = (40 âˆ’ x) = (40 âˆ’ 20) m = 20 m.

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