1-** Check whether the following are quadratic equations:**

(i) (x + 1)^{2} = 2(x – 3)

(ii) x^{2} – 2x = (–2) (3 – x)

Solutions:

(i) Given,

(x + 1)^{2} = 2(x – 3)

By using the formula for (a+b)^{2} = a^{2}+2ab+b^{2}

⇒ x^{2} + 2x + 1 = 2x – 6

⇒ x^{2} + 7 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

Therefore, the given equation is a quadratic equation.

(ii) Given, x^{2} – 2x = (–2) (3 – x)

⇒ x^{2} – 2x = -6 + 2x

⇒ x^{2} – 4x + 6 = 0

Since the above equation is in the form of ax^{2} + bx + c = 0.

Therefore, the given equation is a quadratic equation.

2- **Find two numbers whose sum is 27 and the product is 182.**

Solution:

Let us say, the first number is x and the second number is 27 – x.

Therefore, the product of two numbers

x(27 – x) = 182

⇒ x^{2} – 27x – 182 = 0

⇒ x^{2} – 13x – 14x + 182 = 0

⇒ x(x – 13) -14(x – 13) = 0

⇒ (x – 13)(x -14) = 0

Thus, either, x = -13 = 0 or x – 14 = 0

⇒ x = 13 or x = 14

Therefore, if first number = 13, then second number = 27 – 13 = 14

And if first number = 14, then second number = 27 – 14 = 13

Hence, the numbers are 13 and 14.

3-** The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.**

Solution:

Let us say, the base of the right triangle is x cm.

Given, the altitude of right triangle = (x – 7) cm

From Pythagoras theorem, we know,

Base2 + Altitude2 = Hypotenuse2

∴ x^{2} + (x – 7)^{2} = 132

⇒ x^{2} + x^{2} + 49 – 14x = 169

⇒ 2x^{2} – 14x – 120 = 0

⇒ x^{2} – 7x – 60 = 0

⇒ x^{2} – 12x + 5x – 60 = 0

⇒ x(x – 12) + 5(x – 12) = 0

⇒ (x – 12)(x + 5) = 0

Thus, either x – 12 = 0 or x + 5 = 0,

⇒ x = 12 or x = – 5

Since sides cannot be negative, x can only be 12.

Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.

4- **A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.**

Solution:

Let us say the speed of the train is x km/hr.

Time taken to cover 360 km = 360/x hr.

As per the question given,

⇒ (x + 5)(360/x – 1) = 360

⇒ 360 – x + 1800/x-5 = 360

⇒ x^{2} + 5x – 1800 = 0

⇒ x^{2} + 45x – 40x – 1800 = 0

⇒ x(x + 45) -40(x + 45) = 0

⇒ (x + 45)(x – 40) = 0

⇒ x = 40, -45

As we know, the value of speed cannot be negative.

Therefore, the speed of the train is 40 km/h.

5**- Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.**

Solution:

Let the breadth of the mango grove be l.

Length of the mango grove will be 2l.

Area of mango grove = (2l) (l)= 2l2

2l2 = 800

l2 = 800/2 = 400

l2 – 400 =0

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

a = 1, b = 0, c = 400

As we know, Discriminant = b^{2} – 4ac

=> (0)^{2} – 4 × (1) × ( – 400) = 1600

Here, b^{2} – 4ac > 0

Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed.

l = ±20

As we know, the value of length cannot be negative.

Therefore, breadth of mango grove = 20 m

Length of mango grove = 2 × 20 = 40 m