Find all real number solutions for each equation.
4x^{3}=64x
Lorenzolaji
Answered question
2021-11-17
Find all real number solutions for each equation.
Answer & Explanation
Glenn Cooper
Beginner2021-11-18Added 12 answers
Step 1
Given equation is .
To find all real solution of the given equation.
Solution:
Given equation is .
Solving the given equation.
x(x+4)(x-4)=0 [Using ]
x=0 or (x+4)=0 or x-4=0
x=0 or x=-4 or x=4
Step 2
Therefore, required solution is x={-4,0,4}.
Hence, solution of the given equation is x = {-4, 0 ,4}.
Novembruuo
Beginner2021-11-19Added 26 answers
Calculation:
1)The given polynomial is .
Set the above polynomial as an equation by equate withzero.
That is, .
2)The common monomial factor available in the above equation is 4x.
So, the equation can be written as, .
Therefore, 4x=0 or
3)The above equation can be written as, .
Because, .
Since the above equation has 2 squares and they are subtracted, it represents the difference of two squares pattern. That is, .
4)Now, we can apply the difference of two squares pattern to the above equation. Here, take x=a and 4=b.
Therefore, the equation can be written as follows:
(x+4)(x-4)=0
5)Hence the equation becomes,
x+4=0 or x-4=0.
Therefore, x=-4 or x=4.
Also, we have one more equation from step-2. That is, 4x=0.
Divide both sides of the equation by 4. Hence the equation becomes,
x=0.
So, the solution set is {-4,0,4}.
Final statement:
All the real number solutions for the given polynomial are {-4, 0, 4}.