only number 4, and please include the working. also do it with the substitution method not elimination or any other.
1 answer:
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Answer:
Step-by-step explanation:
There are 6 solutions or zeros here because, according to the Fundamental Theorem of Algebra, the degree of the polynomial dictates how many zeros there are in the polynomial. If we had a 3rd degree polynomial, we would expect to find 3 zeros; if we had a 5th degree polynomial, we would have 5 zeros, etc. The easiest way to factor this is to do it initially by grouping:
then
then
We will factor each set of parenthesis now to get all the zeros. For the first set of parenthesis:
so
so
But since we can't have a negative under the square root, we have to offset it by using the imaginary number i. i-squared = -1, so
x = ±i√5
Those are the first 2 zeros out of 6. Now for the second set of parenthesis:
4x⁴ - 25 = 0. That is the difference between perfect squares, and that factors to this:
(2x² + 5)(2x² - 5)
The first set of parenthesis there:
2x² + 5 = 0 so
2x² = -5 so
x² = -5/2 so
x = ±
Those are the next 2 zeros. We found 4 so far, now we will find the last 2 in the second set of parenthesis above:
so
so
x = ±
In summary, the 6 zeros are as follows:
x = , -
,
,
,
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