10th cbse maths solution for exercise 4.1 part 2

This page 10th cbse maths solution for exercise 4.1 part 2 is going to provide you solution for every problems that you find in the exercise no 4.1

10th cbse maths solution for exercise 4.1 part 2

(vii) (x + 2)³ = 2 x (x² - 1)

Solution:

x³ +  3 (x²) (2) + 3 (x) (2)² + 2³ = 2 x³ - 2 x

x³ +  6 x² + 12 x + 8 = 2 x³ - 2 x

x³ - 2 x³ +  6 x² + 12 x + 2 x + 8 = 0

- x³ + 6 x² + 12 x + 2 x + 8 = 0

This is not exactly matches the general form of quadratic equation. So the given equation is not a quadratic equation.


(vii) x³ - 4 x² - x + 1 = (x - 2)³

Solution:


x³ - 4 x² - x + 1 = x³ -  3 (x²) (2) + 3 (x) (2)² - 2³

x³ - 4 x² - x + 1 = x³ -  6 x² + 12 x - 8

x³- x³ - 4 x² + 6 x² - x - 12 x + 1 + 9 = 0

2 x² - 13 x + 10 = 0

This exactly matches the general form of quadratic equation. So the given equation is a quadratic equation.


(2) Represent the following situations in the form of quadratic equations:

(i) The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Solution:

Area of rectangular plot = 528 m²

let "x" be the breadth of the plot

length of the plot = 2 x + 1

length x breadth = 528

(2 x + 1) x = 528

 2 x² +  x - 528 = 0

This the required quadratic equation.


(ii) the product of two consecutive positive integers is 306. We need to find the integers.

Solution:

Let x and x + 1 are the two consecutive positive integers

 product of these integers = 306

 x (x + 1) = 306

 x² + x = 306

 x² + x - 306 = 0

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