I have been doing pascal triangle and do not know where this question comes from? solve for b
(3+b)to the 4th power = 1
do I take 3 to the 4th + b to the 4th =1 and then subtract 81 so that -81 =b to the fourth which = 3
The 4th row of Pascal's Triangle is
1 4 6 4 1
That is,
(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
So, (3+b)^4 is
3^4 + 4(3^3)b + 6(3^2)b^2 + 4(3^1)b^3 + b^4
= 81 + 108b + 54b^2 + 12b^3 + b^4
That's all very nice, but if your question is to solve for b in
(3+b)^4 = 1
Consider the equation
x^4 = 1
x = ±1,±i
So, you have
3+b = ±1,±i
b = -3±1, -3±i
b = -4,-2,-3+i, -3-i
To solve for b in the equation (3+b)^4 = 1, we need to follow a few steps. Let's break it down:
Step 1: Expand the left side of the equation.
(3+b)^4 can be expanded using the binomial theorem or by using Pascal's triangle.
The fourth row of Pascal's triangle is [1, 4, 6, 4, 1]. Using these coefficients, we can write the expanded form of (3+b)^4 as:
(3^4) + (4 * 3^3 * b) + (6 * 3^2 * b^2) + (4 * 3 * b^3) + (b^4)
Step 2: Simplify the equation.
Now we have: 81 + 108b + 54b^2 + 12b^3 + b^4 = 1
Step 3: Rearrange the equation.
To solve for b, let's move all the terms to one side of the equation:
b^4 + 12b^3 + 54b^2 + 108b + (81 - 1) = 0
b^4 + 12b^3 + 54b^2 + 108b + 80 = 0
Step 4: Solve the equation.
Unfortunately, there isn't a simple method to solve quartic equations (equations with a degree of 4) using basic algebra techniques. To find the solutions for b, you may need to use numerical methods or specialized software.
In this case, it seems like you are assuming a solution for b to be 3. However, it's important to note that this equation can have multiple solutions. Therefore, it's necessary to find all possible solutions by solving the equation in its entirety.