#### Quadratic Equations – Exercise 9.1

- Check whether the following are quadratic equations:

(i) x^{2} – x = 0

(ii) x^{2} = 8

(iii) x^{2} + ^{1}/_{2} x = 0

(iv) 3x – 10 = 0

(v) x^{2} – ^{29}/_{4} x + 5 = 0

(vi) 5 – 6x = ^{2}/_{5}x^{2}

(vii) √2x^{2} + 3x = 0

(viii) √3x = ^{22}/_{13}

(ix)x^{3} – 10x + 74 = 0

(x) x^{2} – y^{2} = 0

2.Simplify the following equations and check whether they are quadratic equations:

(i) x(x + 6) = 0

(ii) (x – 4)(2x – 3) = 0

(iii) (x + 2)(x – 7) = 0

(iv) (x + 1)^{2} = 2(x – 3)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) (x + 2)^{3} = 2x(x^{2} – 1)

3.Represent the following in the form of quadratic equations:

(i) The product of two consecutive integers is 306

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m^{2}

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

#### Quadratic Equations – Exercise 9.1

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

**(i) x ^{2} – x = 0**

Solution:

Given equation is x^{2} – x = 0

The equation is of the form ax^{2} + bx + c = 0, where c = 0

Therefore, x^{2} – x = 0 is quadratic equation.

** (ii) x ^{2} = 8**

Solution:

Given equation is x^{2} = 8

⇒ x^{2} + 0.x – 8 = 0

The equation is of the form ax^{2} + bx + c = 0, where coefficient of x is 0

Therefore, x^{2} – 8 = 0 is quadratic equation.

** (iii) x ^{2} + ^{1}/_{2} x = 0**

Solution:

Given equation is x^{2} +^{1}/_{2} x = 0

The equation is of the form ax^{2} + bx + c = 0, where c = 0

Therefore, x^{2} +^{1}/_{2} x = 0 is quadratic equation.

** (iv) 3x – 10 = 0**

Solution:

Given equation is 3x – 10 = 0

The equation is not of the form ax^{2} + bx + c = 0

Therefore, 3x – 10 = 0 is not a quadratic equation.

** (v) x ^{2} – ^{29}/_{4} x + 5 = 0**

Solution:

Given equation is x^{2} –^{29}/_{4} x + 5 = 0

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

Therefore, x^{2} –^{29}/_{4} x + 5 = 0 is quadratic equation.

** (vi) 5 – 6x = ^{2}/_{5}x^{2}**

Solution:

Given equation is 5 – 6x = ^{2}/_{5}x^{2}

⇒^{ 2}/_{5}x^{2} + 6x – 5 = 0

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

Therefore, 5 – 6x = ^{2}/_{5}x^{2} is quadratic equation.

** (vii) √2x ^{2} + 3x = 0**

Solution:

Given equation is √2x^{2} + 3x = 0

The equation is of the form ax^{2} + bx + c = 0, where c = 0

Therefore, √2x^{2} + 3x = 0 is quadratic equation.

** (viii) √3x = ^{22}/_{13}**

Solution:

Given equation is √3x = ^{22}/_{13}

⇒ √3x – ^{22}/_{13} = 0

The equation is not of the form ax^{2} + bx + c = 0

Therefore, √3x = ^{22}/_{13} is not a quadratic equation.

** (ix)x ^{3} – 10x + 74 = 0**

Solution:

Given equation is x^{3} – 10x + 74 = 0

The equation is not of the form ax^{2} + bx + c = 0

Therefore, x^{3} – 10x + 74 = 0 is not a quadratic equation.

** (x) x ^{2} – y^{2} = 0**

Solution:

Given equation is x^{2} – y^{2} = 0

The equation is of the form ax^{2} + bx + c = 0, where coefficient of x = 0

Therefore, x^{2} – y^{2} = 0 is a quadratic equation.

**2.Simplify the following equations and check whether they are quadratic equations:**

**(i) x(x + 6) = 0**

Solution:

x(x + 6) = 0

x^{2} + 6x = 0

The given equation is x^{2} + 6x = 0

The equation is of the form ax^{2} + bx + c = 0, where c = 0

Therefore, x(x + 6) = 0 is a quadratic equation.

** (ii) (x – 4)(2x – 3) = 0**

Solution:

(x – 4)(2x – 3) = 0

2x^{2} – 3x – 8x + 12 = 0

2x^{2} – 11x + 12 = 0

Therefore, the given equation is (x – 4)(2x – 3) = 0

The equation is of the form ax^{2} + bx + c = 0

Therefore, (x – 4)(2x – 3) = 0 is a quadratic equation.

** (iii) (x + 2)(x – 7) = 0**

Solution:

(x + 2)(x + 7) = 0

x^{2} + 7x + 2x + 14 = 0

x^{2} + 9x + 14 = 0

Therefore, the given equation is (x + 2)(x + 7) = 0

The equation is of the form ax^{2} + bx + c = 0

Therefore, (x + 2)(x + 7) = 0 is a quadratic equation.

** (iv) (x + 1) ^{2} = 2(x – 3)**

Solution:

(x + 1)^{2} = 2(x – 3)

x^{2} + 2x + 1 = 2x – 6

x^{2} + 2x – 2x + 1 – 6 = 0

x^{2} – 5 = 0

Therefore, the given equation is (x + 1)^{2} = 2(x – 3)

The equation is of the form ax^{2} + bx + c = 0

Therefore, (x + 1)^{2} = 2(x – 3) is a quadratic equation.

** (v) (2x – 1)(x – 3) = (x + 5)(x – 1)**

Solution:

(2x – 1)(x – 3) = (x + 5)(x – 1)

2x^{2} – 6x – x + 3 = x^{2} – x + 5x – 5

2x^{2} – 6x – x + 3 – x^{2} + x – 5x + 5 = 0

x^{2} – 12x + 8 = 0

Therefore, the given equation is (2x – 1)(x – 3) = (x + 5)(x – 1)

The equation is of the form ax^{2} + bx + c = 0

Therefore, (2x – 1)(x – 3) = (x + 5)(x – 1) is a quadratic equation.

** (vi) (x + 2) ^{3} = 2x(x^{2} – 1)**

Solution:

(x + 2)^{3} = 2x(x^{2} – 1)

(x + 2)(x^{2} + 2x + 4) = 2x^{3} – 2x

x^{3} + 2x^{2} + 4x + 2x^{2} + 4x + 8 – 2x^{3} – 2x = 0

-x^{3} + 4x^{2} + 6x + 8 = 0

Therefore, the given equation is (x + 2)^{3} = 2x(x^{2} – 1)

The equation is of the form ax^{2} + bx + c = 0

Therefore, (x + 2)^{3} = 2x(x^{2} – 1) is a quadratic equation.

**3.Represent the following in the form of quadratic equations:**

**(i) The product of two consecutive integers is 306**

**(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m ^{2}**

**(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.**

Solution:

(i) The product of two consecutive integers is 306

Let two consecutive numbers be x and x+1.

Then, x(x + 1) = 306

x^{2} + x – 306 = 0

(ii) The length of a rectangular park (in meters) is one more than twice its breadth and its area is 528m^{2}

Area of rectangle = length x breadth

If length of the rectangular park = x m

then, breadth of the rectangular park = (2x + 1) m

(2x + 1)x = 528

2x^{2} + x – 528 = 0

(iii) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance.

Let the speed of train be x km/h

Time taken to travel 480 km = ^{480}/_{x} km/h

In second condition let the speed of the train = (x – 8) km/h

It is also given that the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = (^{480}/_{x} + 3) km/h

Speed x time = distance

(x – 8)(^{480}/_{x} + 3) = 480

480 + 3x – ^{3840}/_{x} – 24 = 0

3x – ^{3840}/_{x} – 24 = 0

3x^{2} – 3840 – 24x = 0

x^{2} – 8x – 1280 = 0

**Next Exercise – Quadratic equations – Exercise 9.2 – Class X**