How do you solve this by factoring?
4p^2 + 121 = 44p
I seriously don't understand and it's frustrating for me. Please help. (:
first you subtract the 44p from both sides to get 4p^2 + 121 - 44p = 0. Then you find two factors of 4p^2. They are 2p and 2p(2p x 2p = 4p^2). Then find two factors of 121 that, when multiplied with the 2p, equals 44p. they are 11 and 11(11 x 11 = 121)
Set the quadratic up like this: (2p - 11)(2p - 11). If you multiply this out, it will equal what you started with!
So this doesn't involve finding pi and sigma then?
x^2-8=7x
To solve the given equation 4p^2 + 121 = 44p by factoring, we need to follow these steps:
Step 1: Move all terms to one side of the equation to make it equal to zero:
4p^2 + 121 - 44p = 0
Step 2: Rearrange the equation in descending order of powers:
4p^2 - 44p + 121 = 0
Step 3: Consider the terms of the quadratic equation in the form of ax^2 + bx + c = 0, where a = 4, b = -44, and c = 121.
Step 4: To factor the equation, we need to find two numbers, let's call them m and n, such that their product (mn) is equal to ac (product of a and c) and their sum (m + n) is equal to b.
In this case:
ac = 4 * 121 = 484
b = -44
We need to find two numbers whose product is 484 and sum is -44.
Step 5: These two numbers are -22 and -22, as (-22) * (-22) = 484 and (-22) + (-22) = -44.
Step 6: Now, rewrite the middle term (-44p) using the two numbers we found in step 5:
4p^2 - 22p - 22p + 121 = 0
Step 7: Group the terms together, factor by grouping, and factor out the common factors from each group:
(4p^2 - 22p) - (22p - 121) = 0
2p(2p - 11) - 1(22p - 121) = 0
Step 8: Factor out the common factors from each group:
2p(2p - 11) - 1(11(2p - 11)) = 0
(2p - 11)(2p - 11) = 0
Step 9: Simplify the equation:
(2p - 11)^2 = 0
Step 10: Now, we have a perfect square equation and we can solve for p:
2p - 11 = 0
Step 11: Solve for p:
2p = 11
p = 11/2
p = 5.5
Therefore, the solution to the equation 4p^2 + 121 = 44p by factoring is p = 5.5.