I need help on doing a math problem for
my algebra class so if u could help i would be really grateful any way please explain.
:
Find two consecutive negative integers with the product of 110
(-11)(-10) = 110
oh wow I thank now I see how to do it
You're welcome.
15000+.002x
To find two consecutive negative integers with a product of 110, we can set up an equation and solve for the integers. Let's call the first integer "x" and the next consecutive integer "x + 1" (since they are negative integers, the next consecutive integer would be smaller than the first).
The equation we can set up is:
x(x + 1) = 110
Now, we can solve this quadratic equation. Expanding the equation, we have:
x^2 + x = 110
Rearranging the equation and setting it equal to zero, we get:
x^2 + x - 110 = 0
This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = 1, and c = -110.
To solve the quadratic equation, you can factor it or use the quadratic formula. In this case, factoring is a suitable approach. We need to find two numbers that multiply to -110 and add up to 1.
Upon factoring, we find that the quadratic equation can be factored as:
(x - 10)(x + 11) = 0
This means that either (x - 10) = 0 or (x + 11) = 0.
For the first case, x - 10 = 0, we solve for x:
x = 10
For the second case, x + 11 = 0, we solve for x:
x = -11
Therefore, the two consecutive negative integers with a product of 110 are -11 and -10.