If difference between the squares of two consecutive numbers is 45. Fi...
Let the number be x,
Then, consecutive number will be (x + 1)
Given that, (x + 1)2 – (x)2 = 45
Using (a + b)2 = a2 + 2ab + b2
⇒ x2 + 1 + 2x – x2 = 45
⇒ 2x = 45 - 1 = 44
⇒ x = 22
∴ The largest number is 23.
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If difference between the squares of two consecutive numbers is 45. Fi...
To find the largest number, we need to solve the given problem step by step.
Let's assume the two consecutive numbers as 'x' and 'x+1'.
The square of the first number is x^2.
The square of the second number is (x+1)^2.
According to the problem, the difference between these squares is 45. So we can form the equation:
(x+1)^2 - x^2 = 45
Expanding the equation:
(x^2 + 2x + 1) - x^2 = 45
Simplifying the equation:
2x + 1 - x^2 = 45
Rearranging the terms:
2x - x^2 - 44 = 0
The equation is now in the form of a quadratic equation. To solve it, we can use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = -1, b = 2, and c = -44. Substituting these values into the formula:
x = (-2 ± √(2^2 - 4(-1)(-44))) / (2(-1))
Simplifying further:
x = (-2 ± √(4 - 176)) / (-2)
x = (-2 ± √(-172)) / (-2)
As we can see, the value inside the square root is negative. This means that there is no real solution to the equation. However, we are looking for the largest number, so we need to consider the positive value of x.
Since there is no real solution, we can conclude that there is no largest number that satisfies the given condition. Therefore, the correct answer is option D) 24.