A boy is 6years younger than his brother and the product of their ages is 135 find their ages
Bb=135
b+6=B
(b+6)b=135
b^2+6b-135=0
now factors of 135, one must end with a 5 (try 15, 9)
His bro. is X years old.
The boy is x-6 years old.
x(x-6) = 135.
x^2--6x-135 = 0,
(x-15)(x+9 = 0,
x-15 = 0, X = 15.
x+9 = 0, X = -9.
X = 15.
x-6 = 9.
make it more explainable
To find the ages of the boy and his brother, we can use a straightforward approach. Let's assume the age of the boy is "x" years, and the age of his brother is "x + 6" years (since the boy is 6 years younger than the brother).
According to the given information, the product of their ages is 135. So, we can set up the equation:
x * (x + 6) = 135
Now, let's solve this equation to find the value of x. We can do this by expanding the equation and rearranging it:
x^2 + 6x = 135
Rearranging the equation:
x^2 + 6x - 135 = 0
Now we have a quadratic equation. To solve it, we can factorize it or use the quadratic formula. In this case, let's factorize it:
(x + 15)(x - 9) = 0
Setting each factor equal to zero:
x + 15 = 0 or x - 9 = 0
x = -15 or x = 9
Since ages cannot be negative, we discard the value x = -15 and consider x = 9 as the age of the boy.
Therefore, the boy is 9 years old and his brother is x + 6 = 9 + 6 = 15 years old.