An integer is 5 more than another integer. Three times the bigger integer is 11 more than the square of the smaller integer. find the two integers
Class 7hindhi
Well, well, well! We have quite the mathematical conundrum here. Let's break it down and see if we can solve it with a sprinkle of humor!
Let's call the smaller integer "x" and the bigger integer "y." According to our problem, we know that y = x + 5.
Now, it's time to dive deeper into the math pool. We have the equation 3y = x^2 + 11.
Let's substitute the value of y from our first equation into the second equation:
3(x + 5) = x^2 + 11.
Now, let's simplify:
3x + 15 = x^2 + 11.
Rearranging this equation, we get:
x^2 - 3x - 4 = 0.
Now, let's summon the great quadratic formula to the rescue!
x = (-b ± √(b^2 - 4ac)) / 2a.
Plugging in our values a = 1, b = -3, and c = -4, we can calculate x.
After some mathematical shenanigans, we find that x equals either 4 or -1.
But hold your horses, which one is it? Let's find out!
If x = 4, we can substitute it back into the first equation to find y:
y = x + 5
y = 4 + 5
y = 9.
If x = -1, we do the same:
y = x + 5
y = -1 + 5
y = 4.
So, we have two possible pairs of integers: (4, 9) and (-1, 4).
Now, isn't math just full of surprises?
Let's call the smaller integer x.
According to the given information, the larger integer is 5 more than x, so we can call it (x + 5).
Three times the bigger integer is 11 more than the square of the smaller integer, so we have the equation:
3(x + 5) = x^2 + 11
Expanding the left side of the equation:
3x + 15 = x^2 + 11
Rearranging the equation:
x^2 - 3x - 4 = 0
This is a quadratic equation that can be factored as:
(x - 4)(x + 1) = 0
Setting each factor equal to zero to solve for x:
x - 4 = 0 -> x = 4
or
x + 1 = 0 -> x = -1
So, the two integers are 4 and -1.
To solve this problem, let's call the smaller integer "x" and the bigger integer "y." We are given two pieces of information:
1) "An integer is 5 more than another integer." In equation form, this can be written as: y = x + 5.
2) "Three times the bigger integer is 11 more than the square of the smaller integer." In equation form, this can be written as: 3y = x^2 + 11.
Now, we have two equations with two variables. We can solve these equations simultaneously to find the values of x and y.
Substituting the value of y from equation 1 into equation 2, we get: 3(x + 5) = x^2 + 11.
Expanding the equation, we have: 3x + 15 = x^2 + 11.
Rearranging the equation, we have: x^2 - 3x - 4 = 0.
Factoring the quadratic equation, we have: (x - 4)(x + 1) = 0.
Setting each factor equal to zero, we have two possible solutions:
1) x - 4 = 0
=> x = 4
2) x + 1 = 0
=> x = -1
So, we have two possible values for x: x = 4 or x = -1.
Substituting these values into equation 1, we can find the corresponding values of y:
1) For x = 4: y = x + 5 = 4 + 5 = 9.
So, the two integers are x = 4 and y = 9.
2) For x = -1: y = x + 5 = -1 + 5 = 4.
So, the two integers are x = -1 and y = 4.
Therefore, the two pairs of integers that satisfy the given conditions are (4, 9) and (-1, 4).
i and i+5
3(i+5) = 11 + i^2
3 i + 15 = 11 + i^2
i^2 - 3 i - 4 = 0
(i-4)(i+1) = 0
i = 4 or i = -1
try 4
then i+5 = 9
is 9 five more than 4, yes
is 27 = 11 + 16 yes
so 4 and 9 work
try -1
i+5 = 4
is 12 = 11 + (-1)^2
yes
so
-1 and 4 work